1003 + Interview Questions in SIMPLE-HARMONIC-MOTION IN GENERAL AWARENESS Page 14 InterviewSolution

651.	A man of mass `60kg` is standing on a platform executing SHM in the vertical plane. The displacement from the mean position varies as `y = 0.5sin(2pift)`. The value of `f`, for which the man will feel weightlessness at the highest point, is (`y` in metre)A. `g//4pi`B. `4pig`C. `sqrt(2g)/(2pi)`D. `2pi sqrt(2g)`
Answer» Correct Answer - C Maximum acceleration = `g` `:. omega ^(2) A = g` `:. (2pi f)^(2) (0.5) = g` `f^(2) = (2g)/(2pi)^(2)` `:. f = sqrt (2g)/(2pi)`.

Discussion

652.	A partilce is executive simple harmonic motion given by `x=5sin(4t-pi/6)` The velocity of the particle when its displacement is 3 units isA. `2pi/3`B. `5pi/6`C. 20D. 16
Answer» Correct Answer - D

Discussion

653.	Two particle execute `SHM` with amplitude `A` and `2A` and angular frequency `omega` and `2omega` repectively. At `t=0` they starts with some initial phase difference. At, `t=(2pi)/(3omega)`. They are in same phase. Their initial phase difference is-A. `(pi)/(3)`B. `(2pi)/(3)`C. `(4pi)/(3)`D. `pi`
Answer» Correct Answer - B::C

Discussion

654.	Two masses M and m are suspended together by massless spring of force constant -k. When the masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillations.A. `(Mg)/k`B. `(mg)/k`C. `((M+m)g)/k`D. `((M-m)g)/k`
Answer» Correct Answer - A

Discussion

655.	Two masses `m_(1)and M_(2)` are suspended together by a massless spring of constant k. When the masses are in equilibrium,`m_(1)`is removed without disturbing the system. Then the angular frequency of oscillation of`m_(2)` is
Answer» Let `l_(1)` and `l_(2)` be the extension produced by masses `m_(1)` and `m_(2)` respectively. Then, `(m_(1) +m_(2))g=k(l_(1)+l_(2))…(i)` After `m_(1)` is removed , `m_(2)g=kl_(2)`…(ii) Eqs. (i) -(ii) gives `m_(1)g=kl_(1)` or `l_(1)=(m_(1)g)/(k)` Now, angular frequency of `m_(2), omega=(2pi)/(T)=2pisqrt((k)/(m_(2)))`

Discussion

656.	Two masses m 1 and m 2 are suspended together by a massless spring of constant K . When the masses are in equilibrium, m 1 is removed without disturbing the system. The amplitude of oscillations is A. `(m_(1)g)/K`B. `(m_(2)g)/K`C. `((m_(1)+m_(2))g)/K`D. `((m_(1)-m_(2))g)/K`
Answer» Correct Answer - A

Discussion

657.	Two masses `m_(1)` and `m_(2)` are suspended together by a massless spring of constant K. When the masses are in equilibrium, `m_(1)` is removed without disturbing the system. Then the angular frequency of oscillation of `m_(2)` is - A. ` sqrt(k/m_(1))`B. ` sqrt(k/m_(2))`C. ` sqrt(k/(m_(1)+m_(2))`D. ` sqrt(k/(m_(1)m_(2))`
Answer» Correct Answer - B

Discussion

658.	Two masses (m_1) and (m_2) are suspended together by a massless spring of spring constant (k). When the masses are in equilibrium, (m_1) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of (m_2). .
Answer» Correct Answer - A::B. When masss (m_(1)) is removed then equilibrum will get disturbed. There will be a restoring forec is the upward direction. The body will undergo S.H.M. now. Let x_(1) be the extension of the spring when (m_(1)_m_(2)) are suspended and x_(2) be the extension of the sppring when `m_(1)` is remved. :. `kx_(1) =(m_(1)+m_(2))g or `(x_91) =(m_(1)+_(m2))g` and `kx_(2) =m_(2)g` or `x_(2) =(m_(2)g)/k` Amplitude of mean position. Restoring force will be or `A =((m_(1_+m_(2))g-m_(2)g)/k =(m_(1))/k` Let at any instant the mass `m_(2)` be having a displacement x from the mean position. Restoring force will be `F=-kx` or `m_(2)a=-kx` rArr `a k/(m_(2)x` Comparing this with `a=-omega^(2)x`, we get `omega^(2)=k/(m^(2) rArr omega=sqrtk/m_(2)`.

Discussion

659.	A particle executes SHM on a straight line path. The amplitude of oscillation is `2cm`. When the displacement of the particle from the mean position is `1cm`, the numerical value of magnitude of acceleration is equal to the mumerical value of velocity. Find the frequency of SHM (in `Hz`).
Answer» Correct Answer - B::C `omega^(2)x = omega sqrt(A^(2) - x^(2)) = sqrt(A^(2) - x^(2))/(x) = 2pif` `:. f = (sqrt (A^(2) - x^(2)))/(2pi x)` Substitute, `A = 2 cm` and `x = 1 cm`

Discussion

660.	The block of mass `m_1` shown in figure is fastened to the spring and the block of mass `m_2` is placed asgainst ilt. A. Find the compression of the spring in the equilibrium position. b.The blocks are pushed a further distance `(2/k)(m_1+m_2)gsintheta` against the spring and released. Find the position where the two blocks separate. c. What is the common speed of blocks at the time of separation?
Answer» Correct Answer - `rarr x=((m_1+m_2)gsintheta)/k` `x_1=2/k(m_1+m_2)gsintheta` `v=sqrt({3/k(m_1+m_2)})gsintheta` `a`. At the equilibrium condition `kx=(m_1+m_2)gsin theta` `rarr x=((m_1+m_2)gsintheta)/k` `b`. `x_1=2/k(m_1+m_2)gsintheta` given When the system is released it will start to make SHM. where , `omega=sqrt(k/(m_1+m_2))` when the blocks lose contact `p=0` So `m_2gsintheta=m_2x_2xxk/(m_1+m_2)` `rarr x_2=((m_1+m_2)gsintheta)/k` so the blocks will lose contact with each other when the springs attain its natural length. `c`.Let the common speed attained you both the block be v `1/2(m_1+m_2)v^2-0` `=1/1k(x_1+x_2)^2-(m_1+m_2)gsintheta(x+x_1)` `[x+x_1=`total compression] `rarr 1/2(m_1+m_2)v^2` `=1/2k(3/k)(m_1+m_2)gsintheta-(m_1+m_2)gsintheta-(x_1+x_2)` `rarr 1/2(m_1_m_2)v^2=` `1/2(m_1+m_2)gsinthetaxx(3/k)(m_1+m_2)gsintheta `rarr `v=sqrt({3/k(m_1+m_2)})gsintheta`

Discussion

661.	In case of a forced vibration, the resonance wave becomes very sharp when theA. applied periodic force is smallB. quality factor is smallC. damping force is smallD. restoring force is small
Answer» Correct Answer - C (c) In resonant vibrations of a body , the frequency of external force applied on the body is equal to its natural frequency . If on increasing and decreasing the frequency of external force from the natural frequency by a factor, the amplitude of vibrations reduces very much. In this case , sharp resonance will take place. But if it reduces by a small factor, then flat resonance will take place. The sharp and flat resonance will depend on damping present in the body execting resonant vibrations. Less the damping , greater will be sharpness.

Discussion

662.	Two blocks connected by a spring rest on a smooth horizontal plane as shown in Fig. A constant force `F` start acting on block `m_2` as shown in the figure. Which of the following statements are not correct?A. length of spring increases continuoulsy, if `m_(1)gtm_(2)`B. block start performing SHM about centre of mass of the system with increasing amplitude.C. blocks start performing SHM about centre of mass of the system which moves rectilinearly with constant accelerationD. acceleration of `m_(2)` is maximum at initial moment of time only
Answer» Correct Answer - C In this case, a constant force F acts on the systme. So, the centre of mass of system moves with constant acceleration. `a_(CM)=(F)/(m_(1)+m_(2))` In the frame of centre of mass, particles execute SHM because force on each particles varies linearly.

Discussion

663.	A simple pendulum is attached to the roof of a lift. If time period of oscillation, when the lift is stationary is T . Then frequency of oscillation, when the lift falls freely, will beA. zeroB. TC. 1/TD. none of these
Answer» Correct Answer - A

Discussion

664.	A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration `alpha`, then the time period is given by `T = 2pisqrt(((I)/(T)))` where g is equal toA. gB. g-aC. g+ aD. `sqrt (g^(2)+a^(2))`
Answer» Correct Answer - D

Discussion

665.	The length of a simple pendulum is 16 cm . It is suspended by the roof of a lift which is moving upwards with an accleration of `6.2 ms^(-2)` . Find the time period of pendulum.
Answer» Given , length of the pendulum (l) =16 cm =0.6m Acceleration of the lift (a) `=6.2 ms^(-2)` `because` Time period (T) `=2pisqrt((l)/((g+a)))=2xx3.14sqrt((0.16)/(9.8+6.2))` `=6.28 xxsqrt((0.16)/(16))` `=6.28xxsqrt((1)/(100))=(6.28)/(10)` `=0.628 s`

Discussion

666.	Assertion : If a pendulum is suspended in a lift and lift accelerates upwards, then its time period will decrease. Reason : Effective value of g will be `g_(e)=g+a`A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - A (a) T=2pidqrt((l)/(g_(e))` Here, `g_(e)=g+a` , if the lift accelerates upwards.

Discussion

667.	A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will beA. TB. `T/4`C. `2T/sqrt5`D. `2Tsqrt5`
Answer» Correct Answer - C

Discussion

668.	A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will beA. `T`B. `T/4`C. `(2T)/(sqrt(5))`D. `2Tsqrt(5)`
Answer» Correct Answer - C `(T_(1))/(T_(2))=sqrt((g+g/4)/g)=(sqrt(5))/2` `T_(2)=(2T)/(sqrt(5))`

Discussion

669.	There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T . If the resultant acceleration becomes `g//4` ,then the new time period of the pendulum isA. 0.8 TB. 0.25 TC. 2 TD. 4 T
Answer» Correct Answer - C

Discussion

670.	The time period of a simple pendulum measured inside a stationary lift is found to be T . If the lift starts accelerating upwards with an acceleration `g //3`, the time period isA. `T/sqrt3`B. `T/3`C. `sqrt3/2T`D. `sqrt3T`
Answer» Correct Answer - C

Discussion

671.	A weakly damped harmonic oscillator of frequency `n_1` is driven by an external periodic force of frequency `n_2`. When the steady state is reached, the frequency of the oscillator will beA. `n_(1)`B. `n_(2)`C. `(n_(1)+n_(2))/(2)`D. `(n_(1)+n_(2))`
Answer» Correct Answer - B (b)With weak damping frequency of system reaches frequency of driving force.

Discussion

672.	Figure shown the kinetic energy `K` of a pendulum versus. its angle `theta` from the vertical. The pendulum bob has mass `0.2kg` The length of the pendulum is equal to `(g = 10m//s^(2))`,
Answer» Correct Answer - 15

Discussion

673.	The bob of a simple pendulum executm simple harmonic motion in water with a period t, while the period of oscillation of the bob is `t_(0)` in air. Negleting frictional force of water and given that the density of the bob is (4//3)xx1000 kg//m^(3). What relationship between t and `t_(0)` is true.
Answer» Correct Answer - 2

Discussion

674.	The bob of a simple pendulum executm simple harmonic motion in water with a period t, while the period of oscillation of the bob is `t_(0)` in air. Negleting frictional force of water and given that the density of the bob is (4//3)xx1000 kg//m^(3). What relationship between t and `t_(0)` is true.A. `t= t_(0)`B. `t=t_(0)//2`C. `t=2t_(0)`D. `t= 4 t_(0)`
Answer» Correct Answer - C

Discussion

675.	The bob of a simple pendulum executm simple harmonic motion in water with a period t, while the period of oscillation of the bob is `t_(0)` in air. Negleting frictional force of water and given that the density of the bob is (4//3)xx1000 kg//m^(3). What relationship between t and `t_(0)` is true.A. (a) `t=2t_(0)`B. (b) `t=t_(0)/2`C. (c ) `t=t_(0)`D. (d) `t=4t_(0)`
Answer» Correct Answer - A (a) `t2pisqrtl/(g_(eff), t_(0)2pisqrtl/g` `mg_(eff)=mg-Bmy-Vxx100xxg` :. G_(eff=g-(100)/(m/v)g=g=(1000)/(4/3xx1000)g=g/4` :. T=2pisqrtl/(g//4)` `t=2t_(0)`.

Discussion

676.	The bob of a simple pendulum executm simple harmonic motion in water with a period t, while the period of oscillation of the bob is `t_(0)` in air. Negleting frictional force of water and given that the density of the bob is (4//3)xx1000 kg//m^(3). What relationship between t and `t_(0)` is true.A. `t=t_(0)`B. `t=t_(0)//2`C. `t=2t_(0)`D. `t=4t_(0)`
Answer» Correct Answer - C

Discussion

677.	What effect occurs on the frequency of a pendulum if it is taken from the earth surface to deep into a mineA. IncreasesB. DecreasesC. First increases then decreaseD. None of these
Answer» Correct Answer - B

Discussion

678.	The bob of a simple pendulum executes simple harmonic motion in water with a period `t`, white the period of oscillation of the bob is `t_(0)` in air. Neglecting frictional force of water and given that the density of the bob is `(4//3) xx 1000 kg//m^(3)`, "Find" `(1)/(t_(0))`.
Answer» Correct Answer - 2

Discussion

679.	The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is CA. Greater than TB. Less than TC. Equal to TD. connot be compared
Answer» Correct Answer - A

Discussion

680.	A mass m = 8kg is attached to a spring as shown in figure and held in positioin so that the spring remains unstretched. The spring constant is 200 N/m. The mass m is then released and begins to undergo small oscillations. The maximum velocity of the mass will be (g=10m/`s^(2))` A. 1 m/sB. 2 m/sC. 4 m/sD. 5 m/s
Answer» Correct Answer - B

Discussion

681.	A weightless spring of length 60 cm and force constant 200 N/m is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass isA. `pi/20 s`B. `pi/10 s`C. `pi/5 s`D. `pi/sqrt(200)s`
Answer» Correct Answer - A

Discussion

682.	A body of mass m falls from a height h on to the pan of a spring balance, figure, The masses of the pan and spring are negligible. The spring constant of the spring is k. The body gets attached to the pan and starts executing S.H.M. in the vertical direction. Find the amplitude and energy of oscillation.
Answer» Correct Answer - `a=(mg)/(k)sqrt(1+(2kh)/((M+m)g)),T=2pisqrt((M+m))//k`

Discussion

683.	A particle is performing S.H.M. Its total energy ie E When the displacement of the particle is half of its amplitude, its K.E. will beA. E/2B. E/4C. 3E/4D. E/8
Answer» Correct Answer - B (b) `U=(1)/(2)momega^(2)x^(2)=(1)/(2)momega^(2)((A^(2))/(4))=(1)/(4)E`

Discussion

684.	A solid ball of mass `m` is made to fail from a height `H` on a pan suspended through a spring of spring constant `K` as shown in figure. If the does not rebound and the pan is massless, then amplitude of oscillation is A. `(mg)/(K)`B. `(mg)/(k)(1 + (2HK)/(mg))^(1//2)`C. `(mg)/(K)+((2HK)/(mg))^(1//2)`D. `(mg)/(K)[1+(1+(2HK)/(mg))^(1//2)]`
Answer» Correct Answer - B

Discussion

685.	A body of mass 2 kg suspended through a vertical spring executes simple harmonic motionof period 4s. If the oscillations are stopped and the body hangs in equillibrium, find the potential energy stored in the spring.
Answer» Correct Answer - `40 J`

Discussion

686.	The displacement of a particle is repersented by the equation `y=3cos((pi)/(4)-2omegat)`. The motion of the particle is(b) (c) (d)A. non-periodicB. periodic but not simple harmonicC. simple harmonic with period `2pi//omega`D. simple harmonic with period `pi//omega`
Answer» Correct Answer - B (b) Given , equation of motion is `y="sin"^(3)omegat=(3 "sin"omegat-4"sin"omegat)//4" "[because "sin"3 theta =3 "sin"theta-4"sin"^(3)theta]` `implies (dy)/(dt)=[(d)/(dt)(3"sin"omegat)-(d)/(dt)(4 "sin"omegat)]//4` `4(dy)/(dt)=3omega"cos"omegat-4xx[3omega"cos"3 omegat]` `implies4xx(d^(2)y)/(dt^(2))=-3omega^(2)"sin"omegat+12 omega^(2)"sin"omegat` `implies(d^(2)y)/(dt^(2))=-(3omega^(2)"sin"omegat+12omega^(2)"sin"3omegat)/(4)` `implies(d^(2)y)/(dt^(2))` is not proportional to y. Hence, motion is not SHM. As the expression is involing sine function ,hence it will be periodic.

Discussion

687.	Two point masses of 3 kg and 6 kg are attached to opposite ends of horizontal spring whose spring constant is `300 Nm^(-1)` as shown in the figure. The natural vibration frequency of the system is approximately A. 4 HzB. 3 HzC. 2 HzD. 1 Hz
Answer» Correct Answer - C (c)Reduced mass of two blocks, `mu=(m_(1)m_(2))/(m_(1)+m_(2))=2 kg` Now, `f=(1)/(2pi)sqrt((k)/(mu))=(1)/(2pi)sqrt((300)/(2))~~2Hz`

Discussion

688.	A spring has a natural length of 50 cm and a force constant of `2.0 xx 10^(3)` N`m^(-1)`. A body of mass 10 kg is suspended from it and the spring is stretched. If the body is pulled down to a length of 58 cm and released, it executes simple harmonic motion. The net force on the body when it is at its lowermost position of its oscillation is (10x) newton. Find value of x. (Take g=10m/`s^(2)`)A. 20 NB. 40 NC. 60 ND. 80 N
Answer» Correct Answer - C (c)At its lowermost point spring is stretched by 8 cm or `8xx10^(-2)m`. `therefore F_("net")=kx-mg` `=(2xx10^(3)xx8xx10^(-2))-10)=60N`

Discussion

689.	Electrons with energy `80 keV` are incident on the tungsten target of an X - rays tube , k- shell electrons of tungsten have `72.5 keV` energy X- rays emitted by the tube contain onlyA. a continous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of `= 0.155 Å`B. a continous X-ray spectrum (Bremsstrahlung) with all wavelengthsC. the characterstic X-ray spectrum of tungstenD. a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of `= 0.155 Å` and the characteristic X-ray spectrum of tungsten.
Answer» Correct Answer - D Minimum wavelength of continuous X-ray spectrum is given by `lambda_("min")` (in Å)`=(12375)/("E(in eV)")` Here, E= energy of incident electron (in eV) = energy corresponding to minimum wavelength `implies lambda_("min")` of X-ray `implies E=80 keV=80 xx10^(3) eV` `:. lambda_("min")` (in Å) `=12375/(80xx10^(3))~~0.155` Also the energy of the incident electrons `(80 keV)` is more than the ionization energy of the K-shell electrons (i.e. 72.5 keV). Therefore, characteristic X-ray spectrum will also be obtained because energy of incident electron is high enough to knock out the electron from K or L-shells.

Discussion

690.	Assertion : All oscillatory motions are necessarily periodic motion but all periodic motion are not oscillatory. Reason : Simple pendulum is an example of oscillatory motion.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false
Answer» Correct Answer - B

Discussion

691.	STATEMENT-1 : An oscillatory motion is always simple harmonic motion. STATEMENT-2 : A simple harmonic motion is always oscillatory motion.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - D

Discussion

692.	STATEMENT-1 : In simple harmonic motionn graph between celocity `(v)` and displacement `(x)` from mean position is elliptical. STATEMENT-2 : Relation between `v` and `x` is given by `(v^(2))/(omega^(2)A^(2))+(X^(2))/(A^(2))=1`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A

Discussion

693.	Two particle are executing SHMs .The equations of their motions are `y_(1)=10"sin"(omegat+(pi)/(4)) " and "y_(2)=5 "sin"(omegat+(sqrt(3)pi)/(4))` What is the ratio of their amplitudes.A. `1:1`B. `2:1`C. `1:2`D. None of these
Answer» Correct Answer - B (b) There is no effect of phase angle on amplitude , Hence `(A_(1))/(A_(2))=(10)/(5)=(2)/(1)`

Discussion

694.	The amplitude of the vibrating particle due to superposition of two `SHMs`, `y_(1)=sin (omega t+(pi)/(3)) and y_(2)=sin omega t` is :A. `1`B. `sqrt2`C. `sqrt3`D. `2`
Answer» Correct Answer - C

Discussion

695.	Function `x=Asin^(2)omegat+Bcos^(2)omegat+Csinomegat cos omegat` represents simple harmonic motion,A. The motion of particle is `SHM` when `A = B`B. The motion of particle is `SHM` when `A = B` and `C = 0`C. If `B = C/2=-A`, then the amplitude of `SHM` is `B//sqrt(2)`.D. If `A=B =C/2`, then the axis of vibration of `SHM` shifts by a distance `B` towards `+x` axis.
Answer» Correct Answer - ACD If ` A = B` then `x = A + C sinomegat cosomegat = A + C/2 sin^(2)omegat rArr "SHM"` If `A = B` & `C = 0` then `x = A rArr ` along a straight line. If `B = C/2 = - A` then `x = B cos^(2)omegat + Bsin2omegat rArr "amplitude" = Bsqrt(2)` If `A = B= C/2` then `x = B + Bsin2omegat rArr` Axis of vibration of SHM shifts by a distance B towards `+ x-` axis.

Discussion

696.	The function sin^(2) (omegat) represents.A. a periodic, butnot simple harmonic, motion with a period `2pi//omega`B. a particle, but not simple harmonic, motion with a period `pi//omega`C. a simple harmonic motion with a period `2pi//omega`D. a simple harmonic motion with a period `pi//omega`
Answer» Correct Answer - B `y = sin^(2)omegat = (I - cos2omegat)/(2) = 1/2 - 1/2 cos2omegat` `rArr` motion is SHM with time period `= (2pi)/(2omega) = (pi)/(omega)`

Discussion

697.	Function `x=Asin^(2)omegat+Bcos^(2)omegat+Csinomegat cos omegat` represents simple harmonic motion,A. for any value of A, B and C (except `C = 0`)B. If `A = -B, C = 2B`, amplitude `= \|Bsqrt(2)\|`C. If `A = B, C = 0`D. If `A = B, C = 2B`, amplitude `= \|B\|`
Answer» Correct Answer - A::B::D For `A = -B` and `C = 2B` `X = B cos 2omegat + B sin 2omegat = sqrt(2)B sin(2omegat+(pi)/(4))` This is equation of SHM of amplitude `sqrt(2)B` If `A = B` and `C = 2B`, then `X = B + B sin 2omegat` This is also equation of SHM about the point `X = B` function oscillates between `X = 0` and `X = 2B` with amplitude `B`.

Discussion

698.	A block P of mass m is placed on horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k, the two blocks are pulled by distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks. A. `kA`B. `kA//2`C. zeroD. `mu_(s) mg`
Answer» Correct Answer - B

Discussion

699.	The function sin^(2) (omegat) represents.A. for any value of `A,B` and `C` (except `C=0`)B. if `A=-B,C=2B`, amplitude=`\|Bsqrt2\|`C. if `A=B, C=0`D. if `A=B,C=2B`, amplitude =`\|B\|`
Answer» Correct Answer - A::B::D

Discussion

700.	A `50 g` mass hangs at the end of a massless spring. When `20 g` more are added to the end of the spring, it stretches `7.0 cm` more. (a) Find the spring constant. (b) If the `20 g` are now removed, what will be the period of the motion ?
Answer» Correct Answer - A::B::D (a) `kx = mg` `:. k = (mg)/(x) = ((20 xx 10^(-3))(10))/(7xx10^(-2))` `= (20)/(7)N//m` (b) `T = 2pi sqrt ((m)/(k))` `= 2pi sqrt ((50 xx10^(-3))/(20//7))` = 0.84 s`

Discussion

Explore topic-wise InterviewSolutions in .

A partilce is executive simple harmonic motion given by `x=5sin(4t-pi/6)` The velocity of the particle when its displacement is 3 units isA. `2pi/3`B. `5pi/6`C. 20D. 16

Two particle execute `SHM` with amplitude `A` and `2A` and angular frequency `omega` and `2omega` repectively. At `t=0` they starts with some initial phase difference. At, `t=(2pi)/(3omega)`. They are in same phase. Their initial phase difference is-A. `(pi)/(3)`B. `(2pi)/(3)`C. `(4pi)/(3)`D. `pi`

Two masses M and m are suspended together by massless spring of force constant -k. When the masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillations.A. `(Mg)/k`B. `(mg)/k`C. `((M+m)g)/k`D. `((M-m)g)/k`

Two masses `m_(1)and M_(2)` are suspended together by a massless spring of constant k. When the masses are in equilibrium,`m_(1)`is removed without disturbing the system. Then the angular frequency of oscillation of`m_(2)` is

Two masses m 1 and m 2 are suspended together by a massless spring of constant K . When the masses are in equilibrium, m 1 is removed without disturbing the system. The amplitude of oscillations is A. `(m_(1)g)/K`B. `(m_(2)g)/K`C. `((m_(1)+m_(2))g)/K`D. `((m_(1)-m_(2))g)/K`

Two masses (m_1) and (m_2) are suspended together by a massless spring of spring constant (k). When the masses are in equilibrium, (m_1) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of (m_2). .

A particle executes SHM on a straight line path. The amplitude of oscillation is `2cm`. When the displacement of the particle from the mean position is `1cm`, the numerical value of magnitude of acceleration is equal to the mumerical value of velocity. Find the frequency of SHM (in `Hz`).

In case of a forced vibration, the resonance wave becomes very sharp when theA. applied periodic force is smallB. quality factor is smallC. damping force is smallD. restoring force is small

A simple pendulum is attached to the roof of a lift. If time period of oscillation, when the lift is stationary is T . Then frequency of oscillation, when the lift falls freely, will beA. zeroB. TC. 1/TD. none of these

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration `alpha`, then the time period is given by `T = 2pisqrt(((I)/(T)))` where g is equal toA. gB. g-aC. g+ aD. `sqrt (g^(2)+a^(2))`

The length of a simple pendulum is 16 cm . It is suspended by the roof of a lift which is moving upwards with an accleration of `6.2 ms^(-2)` . Find the time period of pendulum.

A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will beA. TB. `T/4`C. `2T/sqrt5`D. `2Tsqrt5`

A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will beA. `T`B. `T/4`C. `(2T)/(sqrt(5))`D. `2Tsqrt(5)`

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T . If the resultant acceleration becomes `g//4` ,then the new time period of the pendulum isA. 0.8 TB. 0.25 TC. 2 TD. 4 T

The time period of a simple pendulum measured inside a stationary lift is found to be T . If the lift starts accelerating upwards with an acceleration `g //3`, the time period isA. `T/sqrt3`B. `T/3`C. `sqrt3/2T`D. `sqrt3T`

A weakly damped harmonic oscillator of frequency `n_1` is driven by an external periodic force of frequency `n_2`. When the steady state is reached, the frequency of the oscillator will beA. `n_(1)`B. `n_(2)`C. `(n_(1)+n_(2))/(2)`D. `(n_(1)+n_(2))`

Figure shown the kinetic energy `K` of a pendulum versus. its angle `theta` from the vertical. The pendulum bob has mass `0.2kg` The length of the pendulum is equal to `(g = 10m//s^(2))`,

The bob of a simple pendulum executm simple harmonic motion in water with a period t, while the period of oscillation of the bob is `t_(0)` in air. Negleting frictional force of water and given that the density of the bob is (4//3)xx1000 kg//m^(3). What relationship between t and `t_(0)` is true.

What effect occurs on the frequency of a pendulum if it is taken from the earth surface to deep into a mineA. IncreasesB. DecreasesC. First increases then decreaseD. None of these

The bob of a simple pendulum executes simple harmonic motion in water with a period `t`, white the period of oscillation of the bob is `t_(0)` in air. Neglecting frictional force of water and given that the density of the bob is `(4//3) xx 1000 kg//m^(3)`, "Find" `(1)/(t_(0))`.

The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is CA. Greater than TB. Less than TC. Equal to TD. connot be compared

A particle is performing S.H.M. Its total energy ie E When the displacement of the particle is half of its amplitude, its K.E. will beA. E/2B. E/4C. 3E/4D. E/8

A body of mass 2 kg suspended through a vertical spring executes simple harmonic motionof period 4s. If the oscillations are stopped and the body hangs in equillibrium, find the potential energy stored in the spring.

The displacement of a particle is repersented by the equation `y=3cos((pi)/(4)-2omegat)`. The motion of the particle is(b) (c) (d)A. non-periodicB. periodic but not simple harmonicC. simple harmonic with period `2pi//omega`D. simple harmonic with period `pi//omega`

Two point masses of 3 kg and 6 kg are attached to opposite ends of horizontal spring whose spring constant is `300 Nm^(-1)` as shown in the figure. The natural vibration frequency of the system is approximately A. 4 HzB. 3 HzC. 2 HzD. 1 Hz

Two particle are executing SHMs .The equations of their motions are `y_(1)=10"sin"(omegat+(pi)/(4)) " and "y_(2)=5 "sin"(omegat+(sqrt(3)pi)/(4))` What is the ratio of their amplitudes.A. `1:1`B. `2:1`C. `1:2`D. None of these

The amplitude of the vibrating particle due to superposition of two `SHMs`, `y_(1)=sin (omega t+(pi)/(3)) and y_(2)=sin omega t` is :A. `1`B. `sqrt2`C. `sqrt3`D. `2`

The function sin^(2) (omegat) represents.A. a periodic, butnot simple harmonic, motion with a period `2pi//omega`B. a particle, but not simple harmonic, motion with a period `pi//omega`C. a simple harmonic motion with a period `2pi//omega`D. a simple harmonic motion with a period `pi//omega`

Function `x=Asin^(2)omegat+Bcos^(2)omegat+Csinomegat cos omegat` represents simple harmonic motion,A. for any value of A, B and C (except `C = 0`)B. If `A = -B, C = 2B`, amplitude `= |Bsqrt(2)|`C. If `A = B, C = 0`D. If `A = B, C = 2B`, amplitude `= |B|`

The function sin^(2) (omegat) represents.A. for any value of `A,B` and `C` (except `C=0`)B. if `A=-B,C=2B`, amplitude=`|Bsqrt2|`C. if `A=B, C=0`D. if `A=B,C=2B`, amplitude =`|B|`

A `50 g` mass hangs at the end of a massless spring. When `20 g` more are added to the end of the spring, it stretches `7.0 cm` more. (a) Find the spring constant. (b) If the `20 g` are now removed, what will be the period of the motion ?