956 + Interview Questions in OSCILLATIONS IN GENERAL AWARENESS Page 18 InterviewSolution

851.	A spring-and-block system constitutes a simple harmonic oscillator. To double the frequency of oscillation, the mass of the block must be ….. the initial mass. (A) \(\frac14\) times (B) half (C) double(D) 4 times
Answer» (A) \(\frac14\) times

Discussion

852.	The PE of a simple harmonic oscillator 0.1 s after crossing the mean position is 1/4 of its total energy. Then the period of its oscillation isA. 0.2 sB. 0.3 sC. 0.9 sD. 1.2 s
Answer» Correct Answer - D `PE=(1)/(4)TE" at x"=(A)/(2)` `x=Asinomegat` `(A)/(2)=Asinomegat therefore sinomegat=(1)/(2)` `therefore omegat=sin^(-1)((1)/(2))=(pi)/(6) therefore (2pi)/(T)t=(pi)/(6)` `therefore T=1.2s`

Discussion

853.	Define frequency of simple harmonic motion.
Answer» The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s^-1 or hertz (Hz). Mathematically, frequency is related to time period by f = \(\frac{1}{T}\)

853.

Define frequency of simple harmonic motion.

Answer»

The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s^-1 or hertz (Hz).

Mathematically, frequency is related to time period by f = \(\frac{1}{T}\)

Discussion

854.	A particle executes SHM along x-axis with an amplitude A, time period T with origin as the mean position. At t = 0, if the particle starts in the +ve x-direction from the origin the minimum time in which it will be at `x=-A//2` will beA. `(T)/(12)`B. `(T)/(6)`C. `(T)/(4)`D. `(7T)/(12)`
Answer» Correct Answer - D `-(A)/(2)=Asinomegat,sinomegat=-(1)/(2)impliesomegat=(7pi)/(6)` `t=(7T)/(12)`.

Discussion

855.	A particle of mass `10g` executes a linear `SHM` of amplitude `5 cm` with a pariod of `2s`. Find the `PE` and `KE,(1)/(6) s` after crossing the mean position.
Answer» Given `m = 10g = 10^(-2) kg, T = 2s`, `omega = (2pi)/(T) = (2pi)/(2) = pi rad//s` `A = 5cm = 5xx 10^(-2)m, KE = (1)/(2) mA^(2) omega^(2) cos^(2) omegat` At `t = (1)/(6)s, KE = (1)/(2) xx 10^(-2) xx (5 xx 10^(-2))^(2) (pi^(2)) cos^(2). (pi)/(6)` `= (25 xx 10^(-6))/(2) xx pi^(2) xx ((sqrt(3))/(2))^(2) = 9.25 xx 10^(-5)J` `PE = (1)/(2) mA^(2) omega^(2) sin^(2) omegat` `= (1)/(2) xx 10^(-2) xx (5xx10^(-2))^(2) pi^(2) sin^(2).(pi)/(6)` `= (25 xx 10^(-6))/(2) xx pi^(2) xx ((1)/(2))^(2) = 3.085 xx 10^(-5)J`

Discussion

856.	What is an epoch?
Answer» The phase of a vibrating particle corresponding to time t = 0 is called initial phase or epoch. At, t = ϕ, ϕ = ϕ₀. The constant φ₀ is called initial phase or epoch. It tells about the initial state of motion of the vibrating particle.

856.

What is an epoch?

Answer»

The phase of a vibrating particle corresponding to time t = 0 is called initial phase or epoch. At,

t = ϕ,

ϕ = ϕ₀.

The constant φ₀ is called initial phase or epoch. It tells about the initial state of motion of the vibrating particle.

Discussion

857.	The velocity-time diagram of a harmonic oscillator is shown in the adjoining figure. The frequency of oscillation is A. 25 HzB. 12.25 HzC. 50 HzD. 33.3 Hz
Answer» Correct Answer - A From figure, `T=0.04 therefore n=(1)/(T)=25Hz`

Discussion

858.	A particle of mass 2 g executes SHM with a period of 12 s and amplitude 10 cm. Find the acceleration of the particle and the restoring force on the particle when it is 2 cm from its mean position. Also find the maximum velocity of the particle.
Answer» Data : m = 2g = 2 × 10^-3 kg, T = 12 s, A = 10 cm = 0.1 m, x = ±2 cm = ± 2 × 10^-2 m \(\omega\) = \(\frac{2\pi}T\) = \(\frac{2\times3.142}{12}\) = \(\frac{3.142}{6}\) = 0.5237 rad/s The acceleration of the particle, a = ω² = (0.5237²) (± 2 × 10^-2) = ± 0.2743 × 2 × 10^-2 = ± 5.486 × 10^-3 m/s² The restoring force on the particle at that position, F = ma = ± (2 × 10^-2) (5.486 × 10^-3) = ±1.097 × 10^-5 N The maximum velocity of the particle, v_max = ωA = 0.5237 × 0.1 5.237 × 10^-2 m/s

858.

A particle of mass 2 g executes SHM with a period of 12 s and amplitude 10 cm. Find the acceleration of the particle and the restoring force on the particle when it is 2 cm from its mean position. Also find the maximum velocity of the particle.

Answer»

Data : m = 2g = 2 × 10^-3 kg, T = 12 s,

A = 10 cm = 0.1 m, x = ±2 cm = ± 2 × 10^-2 m

\(\omega\) = \(\frac{2\pi}T\) = \(\frac{2\times3.142}{12}\) = \(\frac{3.142}{6}\) = 0.5237 rad/s

The acceleration of the particle, a = ω² = (0.5237²) (± 2 × 10^-2)

= ± 0.2743 × 2 × 10^-2 = ± 5.486 × 10^-3 m/s²

The restoring force on the particle at that position,

F = ma = ± (2 × 10^-2) (5.486 × 10^-3) = ±1.097 × 10^-5 N

The maximum velocity of the particle, v_max = ωA = 0.5237 × 0.1 5.237 × 10^-2 m/s

Discussion

859.	A particle executes linear SHM with period 12 s. To traverse a distance equal to half its amplitude from the equilibrium position, it takes (A) 6s (B) 4s (C) 2s (D) 1s.
Answer» Correct option is (D) 1s

Discussion

860.	What do you understand by the phase and epoch of an SHM ?
Answer» (1) Phase of simple harmonic motion (SHM) represents the state of oscillation of the particle performing SHM, i.e., it gives the displacement of the particle, its direction of motion from its equilibrium position and the number of oscillations completed. The displacement of a particle in SHM is given by x = A sin (ωt + α). The angle (ωt + α) is called the phase angle or simply the phase of SHM. The SI unit of phase angle is the radian (symbol, rad). (2) Epoch of simple harmonic motion (SHM) represents the initial phase of the particle performing SHM, i.e., it gives the displacement of the particle and its direction of motion at time t = 0. If x₀ is the initial position of the particle, i.e., the position at time t = 0, x₀ = A sin α or α = sin^-1 (x₀/A). The angle α, therefore, determines the initial state of the particle. Hence, the angle α is the epoch or initial phase or phase constant of SHM. [Note : The symbol for the unit radian is rad, not superscripted c.]

860.

What do you understand by the phase and epoch of an SHM ?

Answer»

(1) Phase of simple harmonic motion (SHM) represents the state of oscillation of the particle performing SHM, i.e., it gives the displacement of the particle, its direction of motion from its equilibrium position and the number of oscillations completed.

The displacement of a particle in SHM is given by x = A sin (ωt + α). The angle (ωt + α) is called the phase angle or simply the phase of SHM. The SI unit of phase angle is the radian (symbol, rad).

(2) Epoch of simple harmonic motion (SHM) represents the initial phase of the particle performing SHM, i.e., it gives the displacement of the particle and its direction of motion at time t = 0.

If x₀ is the initial position of the particle, i.e., the position at time t = 0, x₀ = A sin α or α = sin^-1 (x₀/A). The angle α, therefore, determines the initial state of the particle. Hence, the angle α is the epoch or initial phase or phase constant of SHM.

[Note : The symbol for the unit radian is rad, not superscripted c.]

Discussion

861.	A particle of mass 0.5 kg executes SHM. If its period of oscillation is `pi` seconds and total energy is 0.04 J, then the amplitude of oscillation will beA. 40 cmB. 20 cmC. 15 cmD. 10 cm
Answer» Correct Answer - B `T.E.=(1)/(2)momega^(2)A^(2)` `A^(2)=(2T.E.)/(momega^(2))=(2xx0.4)/(0.5xx(4pi^(2))/(pi^(2)))` `A^(2)=0.04 therefore A=2m=20cm`

Discussion

862.	The epoch of a simple harmonic motion represented by `x = sqrt(3)sin omegat + cos omega t m`isA. `30^(@)`B. `40.3^(@)`C. `60^(@)`D. `25^(@)`
Answer» Correct Answer - A `delta=tan^(-1)[(A_(1)sinalpha_(1)+A_(2)sinalpha_(2))/(A_(1)cosalpha_(1)+A_(2)cosalpha_(2))]`

Discussion

863.	Two particles are executing simple harmonic of the same amplitude (A) and frequency `omega` along the x-axis . Their mean position is separated by distance `X_(0)(X_(0)gtA). If the maximum separation between them is (X_(0)+A), the phase difference between their motion is:
Answer» `x_(1) = A sin omegat, x_(2) =Asin (omegat + theta) +x_(0)` `x_(2) -x_(1) = x_(0) +A (sin(omegat +theta)-sin(omegat))` `x_(2) -x_(1) =x_(0) +2A sin ((theta)/(2)) cos (omegat+(theta)/(2))` The distance between the two `SHM` is also oscillating simple harmonically with an amplitude of `x_(0) = 2A sin ((theta)/(2))`. maximum distance between two `SHM` is `x_(0) +A` from the above `x_(0) +2A sin ((theta)/(2)) = x_(0) +A` `sin((theta)/(2)) = (1)/(2), theta = (pi)/(3)`

Discussion

864.	A particle executes `SHM` represented by the equation, `y = 0.02 sin (3.14t+(pi)/(2))m`. Find (i) amplitude (ii) time period (iii) frequecy (iv) epoch (v) maximum velocity and (vi) maximum acceleration.
Answer» Compare the equation `y = 0.02 sin (3.14t+(•)/(2))` with the general form of the equation, `y = A sin (•t+•)` (i) Amplitude `A = 0.01m` (ii) Timw period is given by `T = (2•)/(•)`or `T=(2•)/(3.14) = 2s` (iii) Frequecny `f = (1)/(T) = (1)/(2) Hz = 0.5 Hz` (iv) Epoch `• =(•)/(2) = (3.14)/(2) = 1.57 rad` Epoch is initial phase (v) Maximum velocity `v_(max) = A• = 0.02 xx 3.14 = 0.0628 ms^(-1)` (vi) Maximum acceleration `a_(max) = A•^(2) = 0.02 xx (3.14)^(2) = 0.197 ms^(-2)`

Discussion

865.	The velocity of a particle in `SHM` at the instant when it is `0.6cm` away from the mean position is `4cm//s`. If the amplitude of vibration is `1cm` then its velocity at the instant when it is `0.8cm` away form the mean postion is (in cm//s)A. `2.25`B. `2.5`C. `3.0`D. `3.5`
Answer» Correct Answer - C `V = omega sqrt(A^(2) - x^(2))`,

Discussion

866.	An air chamber of volume V has a neck area of cross section A into which a ball of mass m just fits and can move up and down without any friction, figure. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure volume variations of air to be isothermal.
Answer» Consider an air chamber of volume V with a long nech of uniform area of cross-section A, and a frictioless ball of mass m fitted smoothly in the nech at position , C Fig. the pressure of air below the bal inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD=y There will be decreases in volume and hence increase in pressure of air inside the chamber . The decrease in volume of the air inside the chamber , `DeltaV=Ay` Volumetric strain =Charge in volume/original volume `=(DeltaV)/V=(Ay)/V` `:.` Bulk modulus of electricity E, will be E=stress (or increase in pressure ) /Volumetric strain `=(-p)/(Ay//V)=(-pV)/(Ay)` Here, negative sign shows that the increase in pressure with decrease the volume of air in the chamber. Now, `p=(-EAy)/V` Due to thisi excess pressure, the restoring force acting on the ball is `F=pxxA =(-EAy)/V. A =(-EA^(2))/V y....(i)` Since `F propy` and negative sign show that the force is directed towards equilibrium position. if the applied increased pressure is removed from the ball , the ball will start executing linear SHM in the nech of chamber with C as mean position . In S.H.M , the restoring force, F=-ky comparing (i) and (ii),. we have spring factor, `k=EA^(2)//V` Here, inertia factor =mass of ball =m period , `T=2pisqrt(("inertia factor")/("spring factor"))` `=2pisqrt(m/(EA^(2)//V))=(2pi)/A, sqrt((mV)/E)` `:.` Frequency , v=`1/T =A/(2pi)sqrt(E/(mV))` Note: if the ball osillates in the nech of chamber under isothermal conditions , thru, E=P=picture of air inside the chamber , when ball is at equilibrium position . if the ball oscillate in the neck of chamber under adiabatic conditions, then E=gP. where `g=C_(P)//C_(v)`

Discussion

867.	A spring balance has a scale that reads `0` to `20kg`. The length of the scale is `10cm`. A body suspended from this balance, when displaced and released, oscillates with period of `(pi)/(10)s`. The mass of the body isA. 350.67 NB. 540.11 NC. 311.24 ND. 300.5 N
Answer» Correct Answer - C Time period of the spring mass system, `T = 2pi sqrt((m)/(k))" "...(i)` From spring, mg = kx `rArr" "k=(mg)/(x)=(60 xx 9.8)/(30 xx 10^(-2))=1960 Nm^(-1)" "...(ii)` From Eqs. (i) and (ii), we get `0.8 = 2pi sqrt((m)/(1960))` `rArr" "(0.8 xx 0.8 xx 1960)/(4 pi^(2))=m` `m = 31.81 kg" "rArr" "w = mg` `w = 31.81 xx 9.8 = 311.24 N`

Discussion

868.	A spring balance has a scale that reads `0` to `20kg`. The length of the scale is `10cm`. A body suspended from this balance, when displaced and released, oscillates with period of `(pi)/(10)s`. The mass of the body isA. `2.45 kg`B. `4.9 kg`C. `9.8 kg`D. `19.6 kg`
Answer» Correct Answer - B `T = 2pi sqrt((m)/(K)), Mg = KL, g =9.8 ms^(-2)`

Discussion

869.	A spring balance has a scale that reads from 0 to 50 kg. the length of the scale is 20 cm. a block of mass m is suspended from this balance, when displaced annd released, it oscillates with a period 0.5 s. the value of m is (Take `g=10ms^(-2)`)A. 8 kgB. 12 kgC. 16 kgD. 20 kg
Answer» Correct Answer - C The 20 cm length of the scaler reads upto 50 kg. `thereforeF=mg=(50kg)(10ms^(-2))=500N` and `x=20cm=0.2m` `therefore`Spring constant, `k=(F)/(x)=(500N)/(0.2m)=2500Nm^(-1)` As `T=2pisqrt((m)/(k))` Squaring both sides, we get `T^(2)=(4pi^(2)m)/(k)` `m=(T^(2)k)/(4pi^(2))=((0.5)^(2)xx(2500Nm^(-1)))/(4xx(3.14)^(2))=16kg`

Discussion

870.	A body of mass 20 g connected to a spring of spring constant k, executes simple harmonic motion with a frequency of `(5//pi)` Hz. The value of spring constant isA. `4Nm^(-1)`B. `3Nm^(-1)`C. `2Nm^(-1)`D. `5Nm^(-1)`
Answer» Correct Answer - C Here, `m=20g=20xx10^(-3)kg,upsilon=(5)/(pi)Hz` As, `upsilon=(1)/(2pi)sqrt((k)/(m))` `k=4pi^(2)upsilon^(2)m=4pi^(2)((5)/(pi))^(2)xx20xx10^(-3)=2" N "m^(-1)`

Discussion

871.	A 5 kg collar is attached to a spring of spring constant 500`Nm^(-1)` . It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10 cm and released. Calculate (a) The period of oscillations (b) The maximum speed and (c) maximum acceleration of the collar.
Answer» (a) The periodm of oscillation as given by Eq. (14.21) is, `T = 2pi sqrt((m)/(k))= 2pi sqrt((5.0 kg)/("500 N m"^(-1))` `= (2pi//10)s ` `= 0.63 s` (b) the velocity of the collar executing executting SHM is given by `v (t)=-A omega sin (omega t + phi)` The maximum speed is given by, `v_(m) = A omega` `= 0.1 xx sqrt((k)/(m))` `= 0.1 xx sqrt(("500 N m"^(-1))/(5 kg))` `= 1 ms^(-1)` and it occurs at `x=0` (c) The acceleration of the collar at the displacement `x(t)` from the equilibrium is given by, `a (t) = -omega^(2) x(t)` `= - (k)/(m) x(t)` Therefore, the maximum acceleration is, `a_(max) = omega^(2)A` `= ("500 N m"^(-1))/("5 kg") xx 0.1 m` `= 10 ms^(-2)` and it occurs at the extremitles.

Discussion

872.	In the question number 64, the maximum acceleration of the collar isA. `5ms^(-2)`B. `10ms^(-2)`C. `15ms^(-2)`D. `20ms^(-2)`
Answer» Correct Answer - B Maximum acceleration `a_(max)=omega^(2)A=(k)/(m)A " "(becauseomega=sqrt((k)/(m)))` `=(500" N "m^(-1))/(5kg)xx0.1m=10ms^(-2)`

Discussion

873.	This time period of a particle undergoing SHM is 16 s. It starts motion from the mean position. After 2 s, its velocity is 0.4 `ms^(-1)`. The amplitude isA. 1.44 mB. 0.72 mC. 2.88 mD. 0.36 m
Answer» Correct Answer - A Velocity, `v = r omega cos omega t` `0.4 = r xx (2pi)/(16) xx cos (2 pi)/(16).2 = r xx (2pi)/(16) xx (1)/(sqrt(2))` or `r = (0.4 xx 16 xx sqrt(2))/(2 pi) = (3.2 sqrt(2))/(pi)=1.44 m`

Discussion

874.	A small body attached to one end of a vertically hanging spring is performing (SHM) about is mean position with angular frequency (omega) and amplitude (a). If at a height (y) from the mean position, the body gets detached from the spring, calculate the value of (y) so that the height (H) attained by the mass is maximum. The body does noy interact with the spring during its subsquent motion after detachment. `(a omega^2 gt g)`. .A. `(g)/(omega^(2))`B. `(2g)/(omega^(2))`C. `(3g)/(omega^(2))`D. `(4g)/(omega^(2))`
Answer» Correct Answer - A At distance `y` above then mean position velocity of the block. `v = w sqrt(A^(2) -y^(2))`. After detaching from the spring net downwrd acceleration of the block will be `g`. Therefore, total height attained by the block above the mean position. `h =y + (v^(2))/(2g) =y + (omega^(2)(A^(2) -y^(2)))/(2g)` For `h` to be maximum `(dh)/(dy) = 0` Putting `(dh)/(dy) =0`, we get `y = (g)/(omega^(2))`

Discussion

875.	In figure, `k = 100 N//m, M = 1kg` and `F = 10 N` (a) Find the compression of the spring in the equilibrium position (b) A sharp blow by some external agent imparts a speed of `2 m//s` to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant. (c) Find the time period of the resulting simple harmonic motion. (d) Find the amplitude. (e) Write the potential energy of the spring when the block is at the left estreme. (f) Write the potential energy of the spring when the block is at the right extreme. The answers of (b), (e) and (f) are different. Explain why this does not violate the principle of conservation of energy ?A. `4.5 J`B. `4J`C. `0.5J`D. `2.5J`
Answer» Correct Answer - C Let `delta rarr` compression at mean position `x rarr` amplitude of `SHM :. PE = (1)/(2)k(x -delta)^(2)`

Discussion

876.	In the shown arrangement, both the spring are in their natural lengths. The coefficient of friction between `m_(2)` and `m_(1)` is `mu`. There is no friction between `m_(1)` and the surface. If the blocks are displaced slightly, they together perform simple harmonic motion. Obtain (a) Frequency of such oscillations. (b) The condition if the friction force on clock `m_(2)` is to act in the direction of its displacement from mean position. ( c) If the condition obtained in (b) is met, what can be maximum of their oscillations ?A. `(mum_(2)g(m_(1)+m_(2)))/(m_(1)k_(2)-m_(2)k_(1))`B. `(mum_(1)g(m_(1)+m_(2)))/(m_(2)k_(1)-m_(1)k_(2))`C. `(mu m_(1)g(m_(1)+m_(2)))/(m_(1)k_(1)-m_(2)k_(2))`D. `(mu m_(2)g(m_(1)+m_(2)))/(m_(1)k_(1) -m_(2)k_(2))`
Answer» Correct Answer - A `f = k_(2)A - m_(2) omega^(2)A le mu m_(2)g` `rArr A le (mu m_(2)g)/(k_(2)-m_(2) omega^(2))` and `omega = sqrt(((k_(1)+k_(2))/(m_(1)+m_(2))))`

Discussion

877.	An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it? How dows the spring elongate, when a 1 kg weight is suspended from it? [take `pi=3.14`]
Answer» Correct Answer - 2.45 cm

Discussion

878.	Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is(a) simple harmonic motion.(b) non-periodic motion.(c) periodic motion.(d) periodic but not S.H.M.
Answer» (a) simple harmonic motion. (c) periodic motion.

878.

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is(a) simple harmonic motion.(b) non-periodic motion.(c) periodic motion.(d) periodic but not S.H.M.

Answer»

(a) simple harmonic motion.

(c) periodic motion.

Discussion

879.	Maximum velocity ini SHM is `v_(m)`. The average velocity during motion from one extreme point to the other extreme point will beA. `(4)/(pi)V_(max)`B. `(pi)/(4)V_(max)`C. `(2)/(pi)V_(max)`D. `(pi)/(2)V_(max)`
Answer» Correct Answer - C `v_(max)=Aomega=(2piA)/(T)` and A, Amplitue, T, Period `rArr" "(A)/(T)=(V_(max))/(2pi)` `"Average velocity"=("Total displacement")/("Total time")=(2A)/(T//2)=(4A)/(T)` But `((A)/(T)=(V_(max))/(2pi))` `therefore` Average velocity `= 4 ((V_(max))/(2pi))=(2)/(pi)V_(max)`

Discussion

880.	When a body mass 2.0 kg is suspended by a spring, the spring is stretched if the body is pulled down slightly and released, it oscillated up and down. What forces passes through the mean position? (g = 9.8 N kg-1 )
Answer» Since there is no acceleration in the body at the mean position, hence the resultant force on it will by zero, i.e., the force applied by the spring will be exactly equal to the weight of the body.

Discussion

881.	The maximum velocity and maximum acceleration of a particle per for ming a linear SHM are `alpha and beta` respectively. Then path length of the particle isA. `(alpha^(2))/(beta)`B. `(beta)/(2 alpha^(2))`C. `(2 alpha^(2))/(beta)`D. `(2 beta)/(alpha^(2))`
Answer» Correct Answer - C `alpha = V_(max)=A omega and a_(max)=A omega^(2) = beta` `therefore" "(alpha^(2))/(beta)=(A^(2)omega^(2))/(Aomega^(2))=A` `therefore` Path length `= (2 alpha^(2))/(beta)`

Discussion

882.	A particle is acted simultaneously by mutally perpendicular simple harmonic motion `x=acosomegat` and `y=asinomegat`. The trajectory of motion of the particle will beA. an ellipseB. a parabolaC. a circleD. a straight line
Answer» Correct Answer - C `m = 31.81 kg" "rArr" "w = mg` `w = 31.81 xx 9.8 = 311.24 N` It is an equation of circle.

Discussion

883.	A particle is acted simultaneously by mutally perpendicular simple harmonic motion `x=acosomegat` and `y=asinomegat`. The trajectory of motion of the particle will beA. an ellipesB. a parabolaC. a circleD. a straight line
Answer» Given, `x=acosomegat…..(i)` `y=asinomegat.....(ii)` Squaring and adding Eqs. (i) and (ii), `x^(2)+y^(2)=a^(2)` `(cos^(2)omegat+sin^(2)omegat)` `=a^(2)rArrx^(2)+y^(2)=a^(2)` `[:.cos^(2)omegat+sinomegat=1]` This is the equation, of a circle Clearly, the locus is a circle of constant radius a.

Discussion

884.	A particle is acted simultaneously by mutually perpendicular simple harmonic motions x=a cos`omegat` and `y=asinomegat`. The trajectory of motion of the particle will beA. an ellipseB. a parabolaC. a circleD. a straight line.
Answer» Correct Answer - C Given `x=acosomegat` . . . (i) `y=asinomegat` Squaring and adding (i) and (ii), we get `x^(2)+y^(2)=a^(2)cos^(2)omegat+a^(2)sin^(2)omegat=a^(2)` It is an equation of circle. Thus, trajectory of motion will be circle.

Discussion

885.	What are the two basic characteristics of a simple harmonic motion?
Answer» (a) Acceleration is directly proportional to displacement. (b) Acceleration is directed opposite to displacement.

885.

What are the two basic characteristics of a simple harmonic motion?

Answer»

(a) Acceleration is directly proportional to displacement.

(b) Acceleration is directed opposite to displacement.

Discussion

886.	In simple harmonic motion, the wrong statement isA. velocity of the body is maximum at mean positionB. kinetic energy is minimum at extreme positionC. its acceleration is maximum at extreme position and direction away from mean positionD. its acceleration is minimum at mean position
Answer» Correct Answer - C

Discussion

887.	Which of the following is not characteristics of simple harmonic motion?A. The motion is periodicB. the motion is along a straight line about the mean positionC. The oscillations are responsible for the energy conversion.D. The acceleration of the particle is directed towards the extreme position.
Answer» Correct Answer - D In SHM, the acceleration of the particle is directed towards themean position.

Discussion

888.	The equation of motion of a simple harmonic motion isA. `(d^(2)x)/(dt^(2))=-omega^(2)x`B. `(d^(2)x)/(dt^(2))=-omega^(2)t`C. `(d^(2)x)/(dt^(2))=-omegax`D. `(d^(2)x)/(dt^(2))=-omegat`
Answer» Correct Answer - A In SHM, Acceleration, `a=(d^(2)x)/(dt^(2))=-omega^(2)x`

Discussion

889.	A simple pendulum of length `l` has maximum angular displacement `theta` . Then maximum kinetic energy of a bob of mass `m` isA. `ml//2g`B. `mg//2l`C. `mgl(1-costheta)`D. `mglsin(theta//2)`
Answer» Correct Answer - C

Discussion

890.	What is Relation between SHM and uniform circular motion ?
Answer» Simple harmonic motion is the projection of uniform circular motion upon a diameter of a circle. This circle is called the reference circle and the particles which revolves along it is called reference particle or generating particle.

Discussion

891.	STATEMENT-1`:` Time period of a physical pendulumis independent of mass of the body. STATEMENT-2 `:` Time period ofa torsional pendulum is `2pi sqrt((l)/(k))` where, `l=` moment of inertia and`k=` Torsional constant. STATEMENT-3 `:` S.H.M. is an example of non-uniform motion.A. T.T.TB. F.T.F.C. T.F.F.D. F.F.F
Answer» Correct Answer - 1

Discussion

892.	In ideal simple harmonic motion, the constant quantity isA. amplitudeB. kinetic energyC. potential energyD. force
Answer» Correct Answer - A

Discussion

893.	If the motion of an object repeats itself at regular intervals of time , it is called `"_____"` motion .A. non oscillatory motionB. non periodic motionC. periodic motionD. periodic and oscillatory motion
Answer» Correct Answer - C

Discussion

894.	The displacement equation of a spring block system is given by `y = A sin omegat` in air. It is completely immersed in water. If `A^(1)` and `omega^(1)` be new amplitude and angular freuency thenA. `omega = omega^(1)`B. `A lt A^(1)`C. bothD. none
Answer» Correct Answer - C

Discussion

895.	Two particles execute SHM of the same time period along the same straight lines. They cross each other at the mean position while going in opposite directions. Their phase difference isA. `pi//2`B. `pi`C. `3pi//2`D. `2pi`
Answer» Correct Answer - B

Discussion

896.	Two particles A and B execute SHM on a straight line of path length 2 m starting from the two extreme points simultaneously. If their respective time periods are 1 s and 2 s, the minimum time in which they meet isA. `(1)/(4)s`B. `(1)/(3)s`C. `(1)/(2)s`D. `(2)/(5)s`
Answer» Correct Answer - B `y_(1)=y_(2)` `Asin(omega_(1)t+(pi)/(2))=Asin(omega_(2)t-(pi)/(2))` `cosomega_(1)t=-cosomega_(2)t` `cosomega_(1)t=cos(pi-omega_(2)t)` `omega_(1)t=pi-omega_(2)t` `t=(pi)/(omega_(1)+omega_(2))` `omega_(1)=(2pi)/(T_(1))=2pis,omega_(2)=(2pi)/(2)=pis` `t=(pi)/(2pi+pi)=(1)/(3)s`.

Discussion

897.	What is the minimum condition for a system to execute S.H.N?
Answer» The minimum condition for a body to posses S.H.N. is that it must have elasticity and inertia.

Discussion

898.	What are Persistence of hearing ?
Answer» The impression of sound heard by our ears persist in our mind for 1/10th of a second. So for the formation of distinct beats, frequencies of two sources of sound should be nearly equal (difference of frequencies less than 10).

Discussion

899.	A 3 kg block, attached to a spring, performs linear SHM with the displacement given by x = 2 cos (50t) m. Find the spring constant of the spring.
Answer» Data : m = 3 kg, x = 2 cos (50t) m Comparing the given equation with x = A cos ωt, ω = 50 rad/s ω² = k/m ∴ The spring constant, k = mω²= (3)(50)² = 3 × 2500 = 7500 N/m

899.

A 3 kg block, attached to a spring, performs linear SHM with the displacement given by x = 2 cos (50t) m. Find the spring constant of the spring.

Answer»

Data : m = 3 kg, x = 2 cos (50t) m

Comparing the given equation with x = A cos ωt,

ω = 50 rad/s

ω² = k/m

∴ The spring constant,

k = mω²= (3)(50)²

= 3 × 2500 = 7500 N/m

Discussion

900.	A body of mass m tied to a spring performs SHM with period 2 seconds. If the mass is increased by 3m, what will be the period of SHM ?
Answer» T = 2\(\pi\) \(\sqrt{\frac mk}\) \(\therefore\) \(\frac{T_2}{T_1}\) = \(\sqrt{\frac {m_2}{m_1}}\) = \(\sqrt{\frac {m+3m}{m}}\) = √4 = 2 ∴ T₂ = 2T₁ = 2 × 2 = 4 seconds gives the required period of SHM.

900.

A body of mass m tied to a spring performs SHM with period 2 seconds. If the mass is increased by 3m, what will be the period of SHM ?

Answer»

T = 2\(\pi\) \(\sqrt{\frac mk}\)

\(\therefore\) \(\frac{T_2}{T_1}\) = \(\sqrt{\frac {m_2}{m_1}}\) = \(\sqrt{\frac {m+3m}{m}}\) = √4 = 2

∴ T₂ = 2T₁ = 2 × 2 = 4 seconds gives the required period of SHM.

Discussion

Explore topic-wise InterviewSolutions in .

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A particle is acted simultaneously by mutally perpendicular simple harmonic motion `x=acosomegat` and `y=asinomegat`. The trajectory of motion of the particle will beA. an ellipseB. a parabolaC. a circleD. a straight line

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What are the two basic characteristics of a simple harmonic motion?

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What is Relation between SHM and uniform circular motion ?

In ideal simple harmonic motion, the constant quantity isA. amplitudeB. kinetic energyC. potential energyD. force

If the motion of an object repeats itself at regular intervals of time , it is called `"_____"` motion .A. non oscillatory motionB. non periodic motionC. periodic motionD. periodic and oscillatory motion

The displacement equation of a spring block system is given by `y = A sin omegat` in air. It is completely immersed in water. If `A^(1)` and `omega^(1)` be new amplitude and angular freuency thenA. `omega = omega^(1)`B. `A lt A^(1)`C. bothD. none

Two particles execute SHM of the same time period along the same straight lines. They cross each other at the mean position while going in opposite directions. Their phase difference isA. `pi//2`B. `pi`C. `3pi//2`D. `2pi`

Two particles A and B execute SHM on a straight line of path length 2 m starting from the two extreme points simultaneously. If their respective time periods are 1 s and 2 s, the minimum time in which they meet isA. `(1)/(4)s`B. `(1)/(3)s`C. `(1)/(2)s`D. `(2)/(5)s`

What is the minimum condition for a system to execute S.H.N?

What are Persistence of hearing ?

A 3 kg block, attached to a spring, performs linear SHM with the displacement given by x = 2 cos (50t) m. Find the spring constant of the spring.

A body of mass m tied to a spring performs SHM with period 2 seconds. If the mass is increased by 3m, what will be the period of SHM ?