97 + Interview Questions in LINEAR-AND-ANGULAR-SIMPLE-HARMONIC-MOTION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Two identical blocks A and B, each of mass `m=3kg`, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown in Fig. Another identical blocks C moving velocity `v_0=0.6(m)/(s)` collides with A and sticks to it, as a result, the motion of system takes place in some way Based on this information answer the following questions: Q. After the collision of C and A, the combined body and block B wouldA. oscillate about centre of mass of system and centre of mass is at rest.B. oscillate about centre of mass of system and centre of mass is moving.C. oscillate but about different location other than the centre of mass.D. not oscillate.
Answer» Correct Answer - B As `C` collides with A and sticks to its, the combined mass moves rightwards to compress the spring and hence B moves rightwards due to the spring force, i.e., B accelerates and the combined mass decelerates. The deformation in the spring is changing and the centre of mass of the system continues to move rightwards with contant speed, while both the blocks oscillate about centre of mass of the system. The velocity of the combined mass just after collision is `mv_0=2mv` `v=(v_0)/(2)=0.3(m)/(s)` Velocity of centre of mass of system. `v_(CM)=(2mv+0)/(3m)=(2v)/(3)=(v_0)/(3)=0.2(m)/(s)` time period with which block B and the combined mass oscillate about centre of mass could be computed by using the reduced mass concept. `T=2pisqrt((2m)/(3k))=(pi)/(5sqrt(10))s` Oscillation energy of the system is `E=(2mv^2)/(2)=0.27J` the translational kinetic energy of the centre of mass of system is `E_(CM)=(3mv^2)/(2)=0.18J` The remaining energy is oscillating between kinetic and potential energy during the motion of blocks. For maximum compression in spring either we can use cantre of mass approach or energy approach, here we are using the second method. Oscillation energy `=` Maximum elastic potential energy `0.09=(1)/(2)kx_m^2impliesx_m=3sqrt(10)mm`

Discussion

2.	The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in meters. The period of small oscillation (in sec) isA. `2pi`B. `pi`C. `(pi)/(2)`D. `sqrt2pi`
Answer» Correct Answer - C `F=-(dU)/(dx)=-8sin2x` For small oscillation, `sin2x=2x` i.e., Since `amu-x`, the oscillation are simple harmonic in nature. `T=pisqrt(\|(x)/(a)\|)=2pisqrt((1)/(16))=(pi)/(2)s`

Discussion

3.	A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM isA. `(2a^2)/(3b-a)`B. `(3a^2)/(3a-b)`C. `(2a^2)/(3a-b)`D. `(3a^2)/(3b-a)`
Answer» Correct Answer - C Let the acceleration f, `f=-omega^2x` Therefore distance of the particle from the centre at any time t is given by `x=rcos(omegat)`, where `r` is the amplitude when `t=1s`,`x=r-a` `(r-a)=rcosomega` `cosomega=(r-a)/(r )` .(i) When `t=2`,`x=r-a-b` therefore `r-a-b=cos2omega` `r-a-b=r(2cos^2omega-1)` ..(ii) Substituting the value of `cosomega` from Eq. (i) in Eq. (ii) We get `r-a-b=r[2((r-a)^2)/(r^2)-1]` `=(2(r-a)^2)/(r )-r` `r(3a-b)=2a^2impliesr=(2a^2)/(3a-b)`

Discussion

4.	Assume that a tunnel is dug across the earth (radius=R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of `sqrt((gR)` (b) it is relased from a height R above the tunnel (c ) it is thrown vertically upward along the length of tunnel with a speed of `sqrt(gR).`
Answer» Let `y = A sin (omega t + phi)` At` t = 0` `R = A sin phi` and `(dy)/(dt) = A omega cos theta (omega + phi)` At `t = 0`, the velocity of the partical is downwards direction i.e. `v = - sqrt gR` `(dy)/(dt) = - sqrtgR = A sqrt((g)/( R)) cos phi [as omega = sqrt((g)/(R))]` `A cos phi = - R` `tan phi = - 1 implies phi = (3 pi)/(4)` Squaring and adding Eqs, (i) and (ii), `A = sqrt2 R` `y = sqrt2 sin ((sqrtg)/(R) r + (3 pi)/(4))` At `y = 0, sin ((sqrtg)/(R) r + (3 pi)/(4)) = pi implies t = (pi)/(4) sqrt((R)/(g))` Hence time to sross the tunnel `= (pi)/(2) sqrt((R)/(g))`

Discussion

5.	A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. for small displacement from `x=0`, the motion is simple harmonic.B. if its total mechanical energy is `(k)/(2)`, it has its minimum kinetic energy at the originC. for any finite non zero value of x, there is a force directed away from the originD.
Answer» Correct Answer - A Since `F=-(dU)/(dx)=2kxexp(-x)^2` `F=0` (at equilibrium as `x=0`) U is minimum at `x=0` and `U_(min)=0` `U` is maximum `xrarr+-infty` and `U_(max)=k` The particle would oscillate about `x=0` for small displacement from the origin and it is in stable equalibrium at the origin.

Discussion

6.	In the previous question, the magnitude of velocity of particle at the mentioned instant isA. `(piA)/(T)`B. `(sqrt2piA)/(T)`C. zeroD. `sqrt((7)/(8))xx(2piA)/(T)`
Answer» Correct Answer - B Velocity at any displacement `x` is given by `v=omegasqrt(A^2-x)` So, the required velocity `=(2pi)/(T)sqrt(A^2-(A^2)/(2))=(sqrt2piA)/(T)`

Discussion

7.	In the previous problem, the displacement of the particle from the mean position corresponding to the instant mentioned isA. `(5)/(pi)m`B. `(5sqrt3)/(pi)m`C. `(10sqrt3)/(pi)m`D. `(5sqrt3)/(2pi)m`
Answer» Correct Answer - B From the graph `T=(5-1)=4s` (distance between the two adjacent shown in the figure) and `v_(max)=5(m)/(s)`,`omegaA=5(m)/(s)` `((2pi)/(T))A=5impliesA=(5T)/(2pi)=(5xx4)/(2pi)=(10)/(pi)` m Also `omega=(2pi)/(4)=(pi)/(2)(rad)/(s)` The equation of velocity can be written as `v=5sin((pi)/(2))t(m)/(s)` At extreme position, `v=0`, `sin((pi)/(2))t=0` or `t=2s` Phase of the particle velocity at that insstant corresponding to the above equation `=pi` Therefore, when a phase change of `(pi)/(5)` takes place, the resulting phase`=pi+(pi)/(6)` `v=5sin((pi+(pi)/(6)))=-5sin(pi)/(6)=-5((1)/(2))` `=2.5(m)/(s)` (numerically) `(dy)/(dt)=vimpliesdy=dt` `dy=int5sin((pit)/(2))dt=(10)/(pi)[-cos(((pit)/(2)))]+C` Since at `t=0`, the particle is at the extreme position, there fore at `r=0`,`y=-(10)/(pi)` `-(10)/(pi)=-(10)/(pi)costheta+CimpliesC=0` `y=-(10)/(pi)cos(((pit)/(2)))` Clearly a phase change of `(pi)/(6)` corresponds to a time difference of `(T)/(2pi)((pi)/(6))=(T)/(12)=(4)/(12)=(1)/(3)s` `y=-(10)/(pi)cos((pi)/(6))=-(10)/(pi)((sqrt3)/(2))` `=(5sqrt3)/(pi)m` (numerically) Acceleration `a=(dv)/(dt)=(d)/(dt)(5sin((pit)/(2)))=(5pi)/(2)(cospit)/(2)` `a` at `t=(1)/(3)s=(5pi)/(2)cos((pi)/(6))=(5pisqrt3)/(4)(m)/(s^2)` Maximum displacement `x_(max)=A=(10)/(pi)m` and maximum accceleration, `a_(max)=omega^2A` `=((pi)/(2))^2xx(10)/(pi)=(5pi)/(2)(m)/(s^2)`

Discussion

8.	In the previous question, find maximum velocity and maximum acceleration.
Answer» `nu_(max) = a omega^(2) a = - pi^(2) m//s^(2)` `:. \|a_(max)\| = pi^(2) m//s^(2)`

Discussion

9.	The oscillations represented by curve 1 in the graph are expressed by equation`x=Asinomegat.` The equation for the oscillation represented by curve `2` is expressed asA. `x=2Asin(omegat-(pi)/(2))`B. `x=2Asin(omegat+(pi)/(2))`C. `x=-2Asin(omegat-(pi)/(2))`D. `x=Asin(omegat-(pi)/(2))`
Answer» Correct Answer - A Oscillations represented by curve 2 lags in phase by `(pi)/(2)` and the periods are same. Amplitude of curve 2 is double that of 1. Put `t=0` then `x=2Asin(0-(pi)/(2))=-2A`

Discussion

10.	A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k\|x\|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. Propertional to `(1)/(sqrta)`B. Independent of a.C. Propertional to `sqrta`D. Propertional to `a^(3//2)`
Answer» Correct Answer - A

Discussion

11.	Find the amplitude of the simple harmonic motion obtasined by combining the motions `x_1=(2.0 cm) sinomegat` `and x_2=(2.0cm)sin(omegat+pi/3)`
Answer» The two equation given represent simple hormonic motion along the X-ray with amplitudes `A_(1) = 2.0 cm and A_(2) = 2.0 cm`. The phase different between the two simple harmonic motion is `pi//3`. The resultant imple harmonic motion will have an amplitude. A given by `A = sqrt(A_(1)^(2) + A_(2)^(2) + 2A_(1) , A_(2) cosdelta)` `= sqrt ((2.0 cm)^(2) + (2.0 cm)^(2) + 2 (2.0 cm)^(2) cos) (pi)/(3) = 3.5 cm`

Discussion

12.	The displament of a particular is respresented by the equation ` y =sin ^(3) omegat`. The motion isA. non-periodicB. periodic but not simple harmonicC. simple harmoic with period `2pi//omega`D. simple harmoic with period `pi//omega`
Answer» Correct Answer - B

Discussion

13.	A mass m attached to a spring of spring constant `k` is stretched a distance `x_0` from its equilibrium position and released with no initial velocity. The maximum speed attained by mass in its subsequent motion and the time at which this speed would be attained are, respectively.A. `sqrt((k)(m))x_0`,`pisqrt((m)/(k))`B. `sqrt((k)/(m))(x_0)/(2)`,`(pi)/(2)sqrt((m)/(k))`C. `sqrt((k)/(m))x_0`,`(pi)/(2)sqrt((m)/(k))`D. `sqrt((k)/(m))(x_0)/(2)`,`pisqrt((m)/(k))`
Answer» Correct Answer - C At mean position the speed will be maximum `(kx_0^2)/(2)=(mv^2)/(2)impliesv_(max)=sqrt((k)/(m))x_0` And this is attained at `t=(T)/(4)` Time period of motion is `T=2pisqrt((m)/(k))` So required time is `t=(pi)/(2)sqrt((m)/(k))`

Discussion

14.	A rigid rod of mass m with a ball of mass M attached to the free end is restrained to oscillate in a vertical plane as shown in the figure. Find the natural frequency of oscillation.
Answer» Correct Answer - `(1)/(2 pi) sqrt((3k)/(27 M + 7m))` At equlibrium position deformation of the spring is `x_(0)` `kx_(0) = (1)/(4) = Mg ((3)/(4) l) + mg ((1)/(4))` When the rod is further rotated through an angle `theta` from equlibrium position, the restoring tarque. `tau = - [k(x+ x_(0)) (1)/(4) cos theta - Mg ((3)/(4)) L cos theta] - mg ((L)/(4)) cos theta` `= - [k(x+ x_(0)) (1)/(4) - Mg ((3)/(4)) L - mg ((L)/(4))] cos theta` For small `theta, cos theta ~~ 1` `tau = - (kl)/(4) x` `implies l alpha = - (kl^(2))/(4) theta` `l = m ((3)/(4) L) ^(2) + (mL^(2))/(12) + m ((L)/(4))^(2)` `f = (1)/(2 pi) sqrt((3k)/(27 M + 7m))`

Discussion

15.	A 100 g block is connected to a horizontal massless spring of force constant `25.6(N)/(m)` As shown in Fig. the block is free to oscillate on a horizontal frictionless surface. The block is displaced 3 cm from the equilibrium position and , at `t=0`, it is released from rest at `x=0` It executes simple harmonic motion with the postive `x-direction indecated in Fig. The position time `(x-t)` graph of motion of the block is as shown in Fig. Q. When the block is at position A on the graph, itsA. position and velocity both are negativeB. position is positive and velocity is negativeC. position is negative and velocity is positiveD. position and velocity both the positive
Answer» Correct Answer - B At position A block is in positive region and slope is negative, so velocity is negative.

Discussion

16.	A 100 g block is connected to a horizontal massless spring of force constant `25.6(N)/(m)` As shown in Fig. the block is free to oscillate on a horizontal frictionless surface. The block is displaced 3 cm from the equilibrium position and , at `t=0`, it is released from rest at `x=0` It executes simple harmonic motion with the postive x-direction indecated in Fig. The position time `(x-t)` graph of motion of the block is as shown in Fig. Q. When the block is at position B on the graph its.A. position and velocity both are positive.B. position is positive and velocity is negativeC. position is negative and velocity is positiveD. osition and velocity are negative.
Answer» Correct Answer - C At position B blocks is in negative region and velocity is positive so slope is positive.

Discussion

17.	A soil cylinder of mass `M` and radius `R` is connected to a spring as shown in fig. The cylinder is placed on a rough horizontal surface. All the parts except the cylinder shown in the figure are light. If the cylinder is displaced slightly from its mean position and released, so that it performs pure rolling back and forth about its equilibrium position, determine the time period of oscillation?A. `2pisqrt((M)/(k))`B. `2pisqrt((3M)/(2k))`C. `2pisqrt((3M)/(k))`D. none of these
Answer» Correct Answer - B Let us say in displaced position, the axis of cylinder is at a distance x from its mean position and its velocity of center of mass if x and angular velocity is `omega`. Then, as cylinder is not slipping `v=Romega`. In this position the spring elongates by x. Using energy method we can find frequency of oscillation very easily. Total energy of oscillation is `E=(Iomega^2)/(2)+(Mv^2)/(2)+(kx^2)/(2)` We have `I=(MR^2)/(2)`, so `E=(3)/(4)Mv^2+(kx^2)/(2)` `E=v^2+((2)/(3)(k)/(M))x^2=`constant Comparing with `v^2+omega^2x^2=`constant So, `omega=sqrt((2k)/(3M))impliesT=2pisqrt((3M)/(2k))`

Discussion

18.	A simple pendulum of length 40 cm oscillates with an angular amplitude of 0.04 rad. Find a. the time period b. the linear amplitude of the bob, c. The speed of the bob when the strig makes 0.02 rad with the vertical and d. the angular acceleration when the bob is in moemntary rest. Take `g=10 ms^-2`.
Answer» a. The angular frequency is `omega = sqrt((g)/(l)) = sqrt((10 m//s^(2))/(0.4 m)) = 5 s^(-1)` The time period is `(2 pi)/(omega) = (2 pi)/(5 s^(-1)) = 1.26 s` b. Linear amplitude `= 40 cm xx 0.4 = 1.6 cm` c. Angular speed `theta = omega sqrt(theta_(0)^(2) - theta^(2))` `theta = (5 s^(-1)) sqrt ((0.04)^(2) - (0.02)^(2)) rad = 0.17 rad//s` where speed of the bob at this instant `V = theta A = (40 cm) xx (0.17 s)` `= 6.8 cm//s` d. At momentary rest , the bob is in exterme position. Thus , the angular acceleration `\|alpha\| = omega^(2) theta`. `\|alpha\| = (0.04 rad) (25 s^(2)) = 1 rad//s^(2)`

Discussion

19.	The matallic bob of a simple pendulum has the relative density `rho`. The time period of this pendulum is `T` it the metallic bob is immersed in water the new time period is given byA. `2pisqrt((l)/(ng))`B. `2pisqrt((l)/((1-(1)/(n))g))`C. `2pisqrt((ln)/(g))`D. `2pisqrt((l)/((n-1)g))`
Answer» Correct Answer - B As discussed in theory part `T=2pisqrt((l)/(g(1-(rho)/(sigma))))` `rho=`density of liquid `sigma=`density of bob material Given `rho=(sigma)/(n)` hence period is `2pisqrt((l)/(1-((1)/(n))g))`

Discussion

20.	The period of a simple pendulum whose bob is hollow metallic sphere is `T`. The period is `T_(1)` when the bob is filled with sand, `T_(2)` where it is filled with mercury and `T_(3)` when it is half filled with mercury Which of the following is true? A. `T=T_(1)=T_(2)gtT_(3)`B. `T_(1)=T_(1)=T_(3)gtT`C. `TgtT_(3)gtT_(1)=T_(2)`D. `T=T_(1)=T_(2)ltT_(3)`
Answer» Correct Answer - D

Discussion

21.	A pendulum has time period `T` in air when it is made to oscillate in water it acquired a time period `T = sqrt(2)T` The density of the pendulum bob is equal to (density) of water `= 1)`A. `sqrt(2)`B. 2C. `2sqrt(2)`D. none of these
Answer» Correct Answer - B

Discussion

22.	A cylinderical block of density d stays fully immersed in a beaker filled with two immiscible liquids of different densities `d_1` and `d_2` The block is in equilibrium with half of it in liquid 1 and the other half in liquid 2 as shown in the Fig. If the block is given a displacement downwards released, then neglecting friction study the following statements.A. It executes simple harmonic motion.B. Its motion is periodic but not simple harmonic.C. The frequency of oscillation is independent of the size of the cylinder.D. The displacement of the centre of the cylinder is symmetric about its equilibrium position.
Answer» Correct Answer - A::D since liquid 2 is below liquid 1, liquid 2 is denser than liquid 1. Let area of cross section of the cylindrical block be A and it be displaced downwards by y. then volume of liquid 2 displaced will get increases by `Ay` and that of liquid 1 will get decreased by the same amount `Ay`. Hence net increase in upthrust on the block will be equal to `(Ayd_2g-Ayd_1g)`. This additional upthurust tries to restore the block in original position. It means, the block experiences a restoring force `Ay(d_2-d_1)g`. Since this force is restoring and directly proportional to displacement `y`, it will execute SHM along a vertical line. Hence, option (a) is correct and option (b) is wrong. If mass of the block is equal to `m`, then its acceleration will be equal to ltbr. `(Ayg(d_2-d_1))/(m)`. Since its acceleration depends on mass `m`, frequency of oscillation will depend on size of the cylinder. Hence option (c ) is wrong. If the cylinder is displaced upward through y from equilibrium position, then it will experience a net downward force, equal to calculated above. This shows that its motion will be symmetric about its equilibrium position.

Discussion

23.	Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is `(g=10(m)/(s))`A. `2pi(m)/(s)`B. `pi(m)/(s)`C. `(1)/(pi)(m)/(s)`D. `(1)/(2pi)/(m)/(s)`
Answer» Correct Answer - D Mean position of the particle is `(mg)/(k)` distance below the unstretched position of spring. Therefore aplitude of oscillation `A=(mg)/(k)` `omega=sqrt((k)/(m))=2pif=20pi(f=10Hz)` `(m)/(k)=(1)/(400pi^2)` `v_(max)=Aomega=(g)/(400pi^2)xx20pi=(1)/(2pi)(m)/(s)`

Discussion

24.	If the maximum speed and acceleration of a partical executing SHM is `20 cm//s and 100 cm//s`, find the time period od oscillation.
Answer» Given `\|nu_(max)\| = a omega = 20 cm//s` (i) and `\|f_(max)\| = a omega^(2) = 100 pi cm//s^(2)` (ii) Dividing Eq. (ii) by Eq. (i), we get `omega = 5 pi rad//s` Therefore, time period `T = (2 pi)/(omega) = (2 pi)/(5 pi) = 0.4 s`

Discussion

25.	A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, the veloocity of the particles is 10 cm`//`s. The distance of the particles from the mean position when its speed becomes 5 cm`//`s isA. `sqrt(3)` cmB. `sqrt5` cmC. `2(sqrt3)` cmD. `2(sqrt5)` cm
Answer» Correct Answer - C

Discussion

26.	A partical executes SHM with an amplltude of `10 cm` and friquency `2 Hz, at t = 0`, the partical is at point where potential energy and kinetic energy are same. Find the equation of displacement of partical.
Answer» Let `x = A sin (omega t + phi)` : then `v = (dx)/(dt) = A omega cos (omega t + phi)` `KE = (1)/(2) mv^(2) = (1)/(2) mA^(2) omega^(2) cos ^(2) (omega t + phi)` `(KE)_(max) = (1)/(2) mA^(2) omega^(2) [for cos ^(2) omega t + phi)]` `PE = (1)/(2) mA^(2) - KE = (1)/(2) mA^(2) - mA^(2) omega^(2) cos ^(2) (omega t + phi)` `= (1)/(2) mA^(2) omega^(2) sin ^(2) (omega t + phi)` According to the problem `KE = PE`: therefore `= (1)/(2) mA^(2) omega^(2) sin ^(2) (omega t + phi) = (1)/(2) mA^(2) omega^(2) cos ^(2) (omega t + phi)` `tan^(2) (omega t + phi) = 1 implies tan^(2) (omega t + phi) = tan^(2) (pi)/(4)` `omega t + phi = pi//4 implies phi = pi//4 (t = 0)` `x = A sin (omega t + phi)` Here , `A = 10 cm = 0.1 m` `omega = 2 pi f = 2pi xx 2 = 4 pi rad//s` `phi = pi//4` hence, the equation of SHM is `x = 0.1 sin (4 pi + (pi)/(4))`

Discussion

27.	A partical executing simple harmonic motion has amplitude of `1 m` and time period `2 s At t = 0`, net force on the partical is zero. Find the equation of displacement of the partical.
Answer» In the case SHM , `x = a sin (omega t + phi)` here, `a = 1 m, omega = (2 pi)/(T) = (2 pi)/(2) = pi rad//s` At `t = 0`, net force on the partical position at `t = 0`. `:. x = 0 at t = 0` `:. 0 = a sin (omega xx 0+ phi)` `implies phi = 0` `x = a sin omega t` `x = 1 xx sin omega t` `x = sin omega t`

Discussion

28.	A partical of mass `0.2 kg` executes simple harmonic motion along a path of length `0.2 m` at the rate of `600`oscillations per minute. Assum at `t = 0`. The partical start SHM in positive direction. Find the kinetic potential energies in zoules when the displacement is `x = A//2` where, A stands for the amplitude.
Answer» Given that `f = 600` oscillations//min, `m = 0.2 kg` `A = "Amplitude" = 1//2 xx "Length of path" = 0.1 m = (1)/(10) m` As `v = (1)/(T) = (600)/(60) = 10 Hz impliesomega = 2 pi T = 20 pi rad//s` The magnetitude velocity of a partical performing SHM is given as `v = omega sqrt(A^(2) - x^(2))`. At `x = A//2 = (1)/(20) m`, the velocity at this position `v_(x = A//2) = 20 pi sqrt (((1)/(10)^(2) = (1)/(20)^(2))) = sqrt 3 pi m//s` Hence, kinetic energy at this position `k_(x = A//2) = (1)/(2) mv_(x = A//2) ^(2) = (1)/(2) (0.2) (sqrt 3 pi)^(2) = (3 pi^(2))/(10) J` We can write potential energy at any position `x` as `U = (1)/(2)kx^(2) = (1)/(2) (m omega^(2)) x^(2)` Hence, `U = (1)/(2) (0.2) (20 pi)^(2) ((1)/(20))^(2) = (pi^(2))/(10) J`

Discussion

29.	Two simple pandulum whose lengths are `100cm` and `121cm` are suspended side by side. Then bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will two be in phase again. ?A. 11B. 10C. 21D. 20
Answer» Correct Answer - B

Discussion

30.	The displacement of a particle varies with time according to the relation `y=a sin omega t +b cos omega t `.A. The motion is SHMB. The motion is SHM with amplitude a+bC. The motion is SHM with amplitude `a^(2)+b^(2)`D. The motion is SHM with amplitude `sqrt(a^(2)+b^(2)`
Answer» Correct Answer - A::D

Discussion

31.	Three simple harmonic motions in the same direction having each of amplitude "a" and the same period are superposed. If each differs in phase from the next by `pi//4` thenA. the resultant amplitude is `(1+sqrt(2))`a.B. the phase of the resultant motion relative to first is `90^(@).`C. The energy associated with the resultant motion is `(3+2sqrt2)` times the energy associated with any single motion.D. the resulting motion is not simple harmonic.
Answer» Correct Answer - A::C

Discussion

32.	Two particle `P` and `Q` describe harmonic , motions of same period , same amplitude along the same line about the same equilibrium position `O` when `P` and `Q` are an appose sides of `O` at the same distance from `O` they have the same speed of `1.2m//s` in the same direction when their displacement are the same they have the same speed at `1.6m//s` in apposite direction .The maximum velocity in `m//s` of either particle isA. `2.8`B. `2.5`C. `2.4`D. `2`
Answer» Correct Answer - D

Discussion

33.	Two particle `P` and `Q` describe `S.H.M.` of same amplitude a same frequency `f` along the same straight line .The maximum distance between the two particles is `asqrt(2)` The phase difference between the two particle isA. zeroB. `(pi)/(2)`C. `(pi)/(6)`D. `(pi)/(3)`
Answer» Correct Answer - B

Discussion

34.	The velocity `v` of a particle of mass is moving along a straight line change with time `t` as `(d^(2)v)/(dt^(2)) = - Kv` where `K` is a particle constant which of the following statement is correct?A. the particle does not perform SHM.B. The particle performs SHM with time period `2pisqrt((m)/(K))`C. The particle performs SHM with frequency `(sqrtK)/(2pi)`D. The particle performs SHM with time period `(2pi)/(K)`.
Answer» Correct Answer - C `(d^2v)/(dt)=-kv^2` This equation has standard solution `v=v_0sin(sqrtkt+theta)`, where `omega=sqrtk` hence the particle executes SHM with angular frequency `omega=sqrtk` or frequency `f=(sqrtk)/(2pi)`

Discussion

35.	The equation of displacement of two waves are given as ` y_(1) = 10 sin( 3 pi t + (pi)/(3)) , y_(2) = 5 [ sin 3 pi t + sqrt(3) cos 3 pi t]` Then what is the ratio of their amplitudes
Answer» `y_(2) = 5 sin (3 pi t) + 5 sqrt3 cos (3 pi t)` `A = sqrt(5^(2) + (5sqrt3)^(2)) = 10` `tan phi = sqrt3, phi = pi//3` `y_(2) = 10 sin [3 pi t + (pi)/(3))]`

Discussion

36.	Identify which of the following function represent simple harmonic motion. (i) `Y = Ae^(I omega t)` (ii) `Y = a e^(- omega t)` (iii) `y = a sin^(2) omega t` (iv) `y = a sin omega t + b cos omega t` (v) `y = sin omega t + b cos 2 omega t`
Answer» (i) According to given equation in problem differentiating with respect to time, we get `(dy)/(dt)` `= I A omega e^(i omega t)` differentiating again with respect to time , we get `(d^(2)y)/(dt^(2)) + - omega^(2) A e^(i omega t) = - omega^(2) y [as y = A e^(i omega t)]` Thus we have `(d^(2)y)/(dt^(2)) + omega^(2) y = 0` This is the basic differential equation of SHM. (ii) The function `y = ae^(-omega t)` is not harmonic as it is not expressed in terms of sine and cosine functions, So, it cannot simple harmonic. Moreover, this function is not periodie. (iii) `y = a sin^(2) omega t` The function `y = a sin^(2) omega t` is harmonic. To become simple harmonic `(d^(2)y)/(dt^(2)) prop y` Here, `(dy)/(dt) = 2 a omega sin ^(2) omega t cos omega t` `(d^(2)y)/(dt^(2)) = 2 a omega^(2) [cos^(2) omega t - sin ^(2) omega t] = 2 a omega^(2) [1 - 2 sin^(2) omega t]` `= 2a omega^(2) [1 - (2y)/(a)]` The function is not simple harmonic. (iv) `y = a sin^(2) omega t + b cos omega t` The function `y = a sin omega t` is simple harmonic Because `(dy)/(dt) = omega a cos omega t - omega b sin omega t` `(d^(2)y)/(dt^(2)) = - omega^(2) a sin omega t - omega^(2) b cos omega t implies (d^(2)y)/(dt^(2)) = - omega^(2) y` This is the basic different equation of SHM. ` y = sin omega t + cos 2 omega t` The function `y = sin omega t + cos 2 omega t` is not simple harmonic `(d^(2)y)/(dt^(2)) = - omega^(2) sin omega t - 4 omega^(2) cos 2 omega t = - omega^(2) [sin omega t + 4 cos 2 omega t]` `(d^(2)y)/(dt^(2)) "ne" - omega^(2) y` The function is not simple harmonic.

Discussion

37.	The equation of motion of a particle is `x = a cos (alphat)^(2)` .The motion isA. periodic but not oscillatoryB. periodic and oscillatoryC. oscillatory but not periodicD. neither periodic nor oscillatory
Answer» Correct Answer - C

Discussion

38.	A particle is subjected to two simple harmonic motion in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions. Find the phase difference between the individual motions.
Answer» Let the amplitudes of the individual motion be A each. The resultant ampliotude is also A. If the phase difference between the two motion is `sigma` `A = sqrt(A^(2) + A^(2) + 2 A A cos sigma)` or `= A sqrt(2 (1 + cos sigma) = 2 A cos ((sigma)/(2))` or `cos ((sigma)/(2)) = (1)/(2)` or `sigma = 2 pi//3`

Discussion

39.	The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of `(T)/(6)` it passes its mean position its,A. velocity will be half its maximum velocityB. displacement will be half its amplitudeC. acceleration will be nearly `86%` of its maximumD. `KE=PE`
Answer» Correct Answer - A::C `y=asinomegat=asin((2pit)/(T))` `v=(dy)/(dt)=omegaacos((2pit)/(T))` At `t=(T)/(6)`,`v=omegaacos(((2pi)/(T)(T)/(6)))=(1)/(2)omegaa` or, `v=((v_(max))/(2))` `y=asin((2pi)/(T))xx(T)/(6)=asin(pi)/(3)` It is not half of a. acceleration `=(d^2y)/(dt^2)=(dv)/(dt)=omega^2asin((2pit)/(T))` `=omega^2asin((pi)/(3))=0.86(AC)_(max)` At this instant, `KE=(1)/(2)mv^2=(1)/(2)m((v_(max))/(2))^2=((KE)_(max))/(4)=(1)/(4)(TE)` `PE=TE-KE=(3)/(4)(TE)` i.e., `KEnePE`

Discussion

40.	A spring has natural length `40 cm` and spring constant `500 N//m`. A block of mass `1 kg` is attached at one end of the spring and other end of the spring is attached to a ceiling. The block is relesed from the position, where the spring has length `45 cm`.A. the block will perform SHM of amplitude 5 cm.B. the block will have maximum velocity `30sqrt5.cm//sec.`C. the block will have maximum acceleration 15 `m//s^(2),`D. the minimum potential energy of the spring will be zero.
Answer» Correct Answer - B::C::D

Discussion

41.	The springs shown in the figure are all upstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. A. the amplitude of oscillation is `((K_(2)+K_(3))F)/(K_(1)K_(2)+K_(2)K_(3)+K_(3)K_(1))`B. the oscillation frequency is `(1)/(2pi)[(K_(1)K_(2)+K_(2)K_(3)+K_(3)K_(1))/(M(K_(2)+K_(3)))]^(1//2)`C. the amplitude of oscillation is `(2(K_(2)+K_(3))F)/(K_(1)K_(2)+K_(2)K_(3)+K_(3)K_(1))`D. the oscillation frequency is `(1)/(pi)[(K_(1)K_(2)+K_(2)K_(3)+K_(3)K_(1))/(M(K_(2)+K_(3)))]^(1//2)`
Answer» Correct Answer - A::B

Discussion

42.	A body is in `SHM` with period `T` when oscillated from a freely suspended spring. If this spring is cut in two parts of length ratio `1 : 3 &` again oscillated from the two the two parts separatedly, then the periods are `T_(1) & T_(2)` then find `T_(1)//T_(2)`.
Answer» As we know know, spring constant is inversely is inversely proportional to its natural length . Therefore. `k = ( C)/(l) : k_(1) = ( C)/(l_(1)) and k_(2) = ( C)/(l_(2))` `:. (k_(1))/(k_(2)) = (l_(2))/(l_(1)) = 3` `:. k_(1) = 3 k_(2)` But `(1)/(k) = (1)/(k_(1)) + (1)/(k_(2)) or (1)/(k) = (1)/(3 k_(2)) + (1)/(k_(2)) = (1 + 3)/(3 k_(2))` `k = (3 k_(2))/(4)` `:. k_(2) = (4 k)/(3)` `:. k_(1) = 3 k_(2) - 4k` But `T = 2 pi sqrt((m)/(k))` : `T_(1) = 2 pi sqrt((m)/(k_(1)))` and `T_(2) = 2 pi sqrt((m)/(k_(2)))` `:. (T_(1))/(T_(2)) = sqrt((k_(2))/(k_(1))) = sqrt((4k)/(3 xx 4k))` `(T_(1))/(T_(2)) = sqrt((1)/(3)) = 1: sqrt3`

Discussion

43.	A mass m is suspended from a spring of force constant k and just touches another identical spring fixed to the floor as shown in the fig. The time period of small oscillations isA. `2pisqrt((m)/(k))`B. `pisqrt((m)/(k))+pisqrt((m)/((k)/(2)))`C. `pisqrt((m)/((3k)/(2)))`D. `pisqrt((m)/(k))+pisqrt((m)/(2k))`
Answer» Correct Answer - D When the spring undergoes displacement in the downward direction it completes one-half oscillation while it completes another half oscillation in the upward direction. The total time period is `T=pisqrt((m)/(k))+pisqrt((m)/(2k))`

Discussion

44.	A particle is in linear simple harmonic motion between two points `A` and `B 10cm` apart (figure). Take the direction form `A` to `B` as positive direction and choose the correct statements. `AO =OB = 5cm` `BC =8cm` A. The sign of velocity , acceleration and force on the particle when it is 3 cm away from A going towards B are positiveB. The sign of velocity of the particle at C given towards B is negativeC. The sign of velocity , accelecration and force on the particle when it sis 4 cm away from B going towards A are negativeD. The sign of acceleration and from ao nt particle when it si at ponts B is negative
Answer» Correct Answer - A::C::D

Discussion

45.	A body of mass `m` is attached by an inelastic string to a suspended spring of spring constant `k`. Both the string and the spring have negligible mass and the string is inextensible and of length `L`. Initially, the mass `m` is at rest. The largest amplitude `A_(max)`, for which the string will remain taut throughout the motion isA. `(mg)/(2k)`B. `(mg)/(k)`C. `(2mg)/(3k)`D. `L`
Answer» Correct Answer - B In the position of equilibrium when the tension in the string `=mg` the extension in the spring is `x_0=(mg)/(k)`. If the amplitude exceeds this value, then in its upward motion, the body will rise to a height at which the string will become loose and `T=0` Hence maximum amplitude is `x_m=(mg)/(k)`. It also follows from the fact that for `Tge0` the maximum downwards acceleration that the body can have is `g`. Since `f(t)=(g-T(t))/(m)`, `f(t)leg`. Hence maximum amplitude is `A_m=(g)/(omega^2)=(mg)/(k)`

Discussion

46.	A small block of mass m is fixed at upper end of a massive vertical spring of spring constant `k=(2mg)/(L)` and natural length `10L` The lower end of spring is free and is at a height L from fixed horizontal floor as shown. The spring is initially unstressed and the spring block system is released from rest in the shown position. Q. Till the blocks reaches its lowest position for the first time, the time duration for which the spring remains compressed isA. `pisqrt((L)/(2g))+sqrt((L)/(4g))sin^-1(1)/(3)`B. `(pi)/(4)sqrt((L)/(g))+sqrt((L)/(4g))sin^-1(1)/(3)`C. `pisqrt((L)/(2g))+sqrt((L)/(4g))sin^-1(2)/(3)`D. `(pi)/(2)sqrt((L)/(2g))+sqrt((L)/(4g))sin^-1(2)/(3)`
Answer» Correct Answer - B `V_(max)=(3)/(2)sqrt(gL)` and `omega=sqrt((k)/(m))=2sqrt((g)/(L))` `A=(V_(max))/(omega)=(3)/(4)L` Hence time taken t, from start of compression till the block reaches mean position is given by `x=Asinomegat_0` where `x=(L)/(4)` `t_0=sqrt((L)/(4g))sin^-1((1)/(3))` Time taken by the block to reach from mean position to bottom most position is `(2pi)/(4omega)=(pi)/(4)sqrt((L)/(g))` Hence the required time `=(pi)/(4)sqrt((L)/(g))+sqrt((L)/(4g))sin^-1(1)/(3)`

Discussion

47.	A small block of mass m is fixed at upper end of a massive vertical spring of spring constant `k=(2mg)/(L)` and natural length `10L` The lower end of spring is free and is at a height L from fixed horizontal floor as shown. The spring is initially unstressed and the spring block system is released from rest in the shown position. Q. As the block is coming down, the maximum speed attained by the block isA. `sqrt(gL)`B. `sqrt(3gL)`C. `(3)/(2)sqrt(gL)`D. `sqrt((3)/(2)gL)`
Answer» Correct Answer - C At the instant block is in equilibrium position its speed is maximum and compression in spring is x given by `kx=mg` (i) From conservation energy, `mg(L+x)=(1)/(2)kx^2+(1)/(2)mv_(max)^2` .(ii) From Eqs (i) and (ii) we get `v_(max)=(3)/(2)sqrt(gL)`

Discussion

48.	The acceleration-displacement `(a-X)` graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscillation.
Answer» In SHM , `alpha = - omega^(2) x` `beta = - omega^(2) = (- alpha)` `:. beta = omega^(2) = alpha` (from graph) and `omega = sqrt((beta)/(alpha))` or `2 pi f = sqrt ((beta)/(alpha))` or `f = (1)/(2 pi) sqrt((beta)/(alpha))`

Discussion

49.	A graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion isA. B. C. D.
Answer» Correct Answer - D `x=Asinomegat` `v=(dx)/(dt)=Aomegat=omegasqrt(A^2-x^2)` `a=(dv)/(dt)=-Aomega^2sinomegat` `=(dv)/(dt)=-omega^2x` But `x=-(1)/(omega^2)` `v=omegasqrt(A^2-(a^2)/(omega^4))` or `v^2=omega^2(A^2-(a^2)/(omega^4))`

Discussion

50.	A linear harmonic oscillator has a total mechanical energy of `200 J`. Potential energy of it at mean position is `50J`. Find (i) the maximum kinetic energy, (ii)the minimum potential energy,(iii) the potential energy at extreme positions.
Answer» At mean position , potential energy is minimum and kinetic energy is maximum . Hence. `U_(min) = 50 J` (at mean position) And ` K_(max) = E - U_(min) = 200 - 50` `K_(max) = 150 J` (at mean position) At exterme positions, kinetic energy is zero and potential energy is maximum `:. U_(max) = E` `= 200 J` (at exterme position)

Discussion

Explore topic-wise InterviewSolutions in .

The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in meters. The period of small oscillation (in sec) isA. `2pi`B. `pi`C. `(pi)/(2)`D. `sqrt2pi`

A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same direction. The amplitude of the SHM isA. `(2a^2)/(3b-a)`B. `(3a^2)/(3a-b)`C. `(2a^2)/(3a-b)`D. `(3a^2)/(3b-a)`

In the previous question, the magnitude of velocity of particle at the mentioned instant isA. `(piA)/(T)`B. `(sqrt2piA)/(T)`C. zeroD. `sqrt((7)/(8))xx(2piA)/(T)`

In the previous problem, the displacement of the particle from the mean position corresponding to the instant mentioned isA. `(5)/(pi)m`B. `(5sqrt3)/(pi)m`C. `(10sqrt3)/(pi)m`D. `(5sqrt3)/(2pi)m`

In the previous question, find maximum velocity and maximum acceleration.

The oscillations represented by curve 1 in the graph are expressed by equation`x=Asinomegat.` The equation for the oscillation represented by curve `2` is expressed asA. `x=2Asin(omegat-(pi)/(2))`B. `x=2Asin(omegat+(pi)/(2))`C. `x=-2Asin(omegat-(pi)/(2))`D. `x=Asin(omegat-(pi)/(2))`

Find the amplitude of the simple harmonic motion obtasined by combining the motions `x_1=(2.0 cm) sinomegat` `and x_2=(2.0cm)sin(omegat+pi/3)`

The displament of a particular is respresented by the equation ` y =sin ^(3) omegat`. The motion isA. non-periodicB. periodic but not simple harmonicC. simple harmoic with period `2pi//omega`D. simple harmoic with period `pi//omega`

A rigid rod of mass m with a ball of mass M attached to the free end is restrained to oscillate in a vertical plane as shown in the figure. Find the natural frequency of oscillation.

The matallic bob of a simple pendulum has the relative density `rho`. The time period of this pendulum is `T` it the metallic bob is immersed in water the new time period is given byA. `2pisqrt((l)/(ng))`B. `2pisqrt((l)/((1-(1)/(n))g))`C. `2pisqrt((ln)/(g))`D. `2pisqrt((l)/((n-1)g))`

A pendulum has time period `T` in air when it is made to oscillate in water it acquired a time period `T = sqrt(2)T` The density of the pendulum bob is equal to (density) of water `= 1)`A. `sqrt(2)`B. 2C. `2sqrt(2)`D. none of these

Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is `(g=10(m)/(s))`A. `2pi(m)/(s)`B. `pi(m)/(s)`C. `(1)/(pi)(m)/(s)`D. `(1)/(2pi)/(m)/(s)`

If the maximum speed and acceleration of a partical executing SHM is `20 cm//s and 100 cm//s`, find the time period od oscillation.

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, the veloocity of the particles is 10 cm`//`s. The distance of the particles from the mean position when its speed becomes 5 cm`//`s isA. `sqrt(3)` cmB. `sqrt5` cmC. `2(sqrt3)` cmD. `2(sqrt5)` cm

A partical executes SHM with an amplltude of `10 cm` and friquency `2 Hz, at t = 0`, the partical is at point where potential energy and kinetic energy are same. Find the equation of displacement of partical.

A partical executing simple harmonic motion has amplitude of `1 m` and time period `2 s At t = 0`, net force on the partical is zero. Find the equation of displacement of the partical.

Two simple pandulum whose lengths are `100cm` and `121cm` are suspended side by side. Then bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will two be in phase again. ?A. 11B. 10C. 21D. 20

The displacement of a particle varies with time according to the relation `y=a sin omega t +b cos omega t `.A. The motion is SHMB. The motion is SHM with amplitude a+bC. The motion is SHM with amplitude `a^(2)+b^(2)`D. The motion is SHM with amplitude `sqrt(a^(2)+b^(2)`

Two particle `P` and `Q` describe `S.H.M.` of same amplitude a same frequency `f` along the same straight line .The maximum distance between the two particles is `asqrt(2)` The phase difference between the two particle isA. zeroB. `(pi)/(2)`C. `(pi)/(6)`D. `(pi)/(3)`

The equation of displacement of two waves are given as ` y_(1) = 10 sin( 3 pi t + (pi)/(3)) , y_(2) = 5 [ sin 3 pi t + sqrt(3) cos 3 pi t]` Then what is the ratio of their amplitudes

Identify which of the following function represent simple harmonic motion. (i) `Y = Ae^(I omega t)` (ii) `Y = a e^(- omega t)` (iii) `y = a sin^(2) omega t` (iv) `y = a sin omega t + b cos omega t` (v) `y = sin omega t + b cos 2 omega t`

The equation of motion of a particle is `x = a cos (alphat)^(2)` .The motion isA. periodic but not oscillatoryB. periodic and oscillatoryC. oscillatory but not periodicD. neither periodic nor oscillatory

A particle is subjected to two simple harmonic motion in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions. Find the phase difference between the individual motions.

A body is in `SHM` with period `T` when oscillated from a freely suspended spring. If this spring is cut in two parts of length ratio `1 : 3 &` again oscillated from the two the two parts separatedly, then the periods are `T_(1) & T_(2)` then find `T_(1)//T_(2)`.

The acceleration-displacement `(a-X)` graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscillation.

A graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion isA. B. C. D.

A linear harmonic oscillator has a total mechanical energy of `200 J`. Potential energy of it at mean position is `50J`. Find (i) the maximum kinetic energy, (ii)the minimum potential energy,(iii) the potential energy at extreme positions.