956 + Interview Questions in OSCILLATIONS IN GENERAL AWARENESS Page 13 InterviewSolution

601.	A body executes `SHM` has its velocity `10cm//s` and `7cm//s` when its displacements from mean positions are `3cm` and `4cm` respectively. The length of path is nearlyA. `10cm`B. `9.5cm`C. `4cm`D. `11.36cm`
Answer» Correct Answer - B `A = sqrt((v_(1)^(2)y_(2)^(2)-v_(2)^(2)y_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`, length of the path `=2A`

Discussion

602.	In a simple pendulum if iron sphere is replaced by a wooden sphere of same mass its time period of oscillation willA. increasesB. decreasesC. remains sameD. It does not oscillate
Answer» Correct Answer - A

Discussion

603.	In case of a forced vibration, the resonance wave becomes very sharp when theA. restroring force is smallB. damping force is smallC. quality factor is smallD. applied periodic force is small.
Answer» Correct Answer - B Lesser the damping force more sharp is the resonance peak.

Discussion

604.	A body executing S.H.M.along a straight line has a velocity of `3 ms^(-1)` when it is at a distance of 4m from its mean position and`4ms^(-1)` when it is at a distance of 3m from its mean position.Its angular frequency and amplitude areA. `2 rad s^(-1) & 5m`B. `1 rad s^(-1) & 10 m`C. ` 2 rad s^(-1)& 10 m`D. ` 1 rad s^(-1) & 5m`
Answer» Correct Answer - 4

Discussion

605.	A particle performing `SHM` with a frequency of `5Hz` and amplitude `2cm` is initially at position exterme position. The equation for its displacement (in metre) isA. `x = 0.02 sin 10 pit`B. `x = 0.02 sin 5 pit`C. `x = 0.02 cos 10 pit`D. `x = 0.02 cos 5pi t`
Answer» Correct Answer - C `x - A cos omegat`

Discussion

606.	A particle executing `SHM` from exterme position towards centre is observed to be at distances `x_(1),x_(2)` and `x_(3)` from the centre at the end of three successive seconds. The period of `SHM` is.
Answer» `x_(1) = A cos omega, x_(2) = A cos 2 omega ,x_(3) =A cos 3 omega` `x_(1) +x_(3) =A [cos omega + cos 3 omega]` from, `cos A +cos B = 2 cos ((A+B)/(2)) cos((A-B)/(2))` `x_(1) +x_(3) = A[2 cos omega cos 2 omega]` `(x_(1)+x_(3))/(2x_(2)) = cos omega rArr omega = cos^(-1) [(x_(1)+x_(3))/(2x_(2))]` `T = (2pi)/(cos^(-1)[(x_(1)+x_(2))/(2x_(2))])`

Discussion

607.	The frequency of a particle performing `SHM` is `12Hz`. Its amplitude is `4cm`. Its initial displacement is `2cm` towards positive exterme positions. Its equation for displacement isA. `x = 0.04 cos (24pi t+(pi)/(6))m`B. `x = 0.04sin (24 pit)m`C. `x = 0.04 sin(24pit+(pi)/(6))m`D. `x = 0.04 cos (24 pit)m`
Answer» Correct Answer - C `x = A sin (omegat +phi)`

Discussion

608.	`A` mass `M` is suspended from a light spring. An additional mass `m` added to it displaces the spring further by distance `x` then its time period isA. `T = 2pi sqrt((mg)/(x(M+m)))`B. `T = 2pi sqrt(((M+m)x)/(mg))`C. `T = 2pi sqrt(((M+m)x)/(Mg))`D. `T = 2pi sqrt((Mg)/(mg))`
Answer» Correct Answer - B

Discussion

609.	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 continuousky if `m_(1) gt m_(2)`B. blocks start performing `SHM` about centre of mass of the system with increasing amplitudeC. blocks start performing `SHM` about centre of mass of the system which moved rectilineraly with constant accelerationD. acceleration of `m_(2)` is maximum at initial moment of time only
Answer» Correct Answer - C

Discussion

610.	In case of a forced vibration, the resonance wave becomes very sharp when theA. applied periodic force is smaalB. quality factor is smallC. damping force is smallD. restoring force is small
Answer» Correct Answer - C 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 executing resonant vibrations. Less the damping, greater will be sharpness.

Discussion

611.	A particle executing SHM completes 1200 oscillations per minute and passes through the mean position with a velocity of `31.4ms^(-1)`. Determine the maximum displacement of the particle from the mean position also obtain the displacement equation of the aprticle if its displacemet be zero at the instant t=0.
Answer» Correct Answer - A=0.025m, y=0.025 sin `(40pit)` metre

Discussion

612.	A body executes `SHM`, such that its velocity at the mean position is `1ms^(-1)` and acceleration at exterme position is `1.57 ms^(-2)`. Calculate the amplitude and the time period of oscillation.
Answer» `(a_(max))/(v_(max)) = (A•^(2))/(A•) = (1.57)/(1) rArr• = 1.57 rad` `:.` Time period `T = (2•)/(1.57) = (2(3.14))/(1.57) = 4s`. but `A• = 1`, i.e., `A (1.57) = 1 or A = (1)/(1.57)` `:.` Amplitude `A = 0.637m`.

Discussion

613.	A clock `S` is based on oscillations of a spring and clock `P` is based on pendulum motion, both clocks run at the same rate on Earth. On a planet having the same mass, but twice the radius that of the earthA. `P` will run slower than `S`B. `P` will run faster than `S`C. They will both run at the same rate as on EarthD. Both do not dunction.
Answer» Correct Answer - A

Discussion

614.	A clock `S` is based on oscillations of a spring and clock `P` is based on pendulum motion, both clocks run at the same rate on Earth. On a planet having the same mass, but twice the radius that of the earthA. S will run faster than PB. P will run faster than SC. both will run at the same rate as on the earthD. both will run at the same rate which will be different from that on the earth
Answer» Correct Answer - B Acceleration due to gravity, `g = (GM)/(R^(2))=(Gxx(4)/(3)pi R^(3)rho)/(R^(2))=(4)/(3)pi G rho R` or `g prop R` For pendulum clock, g will increase on the plane, so time period will decrease. But for spring clock, it will not change. Hence, P will run faster than S.

Discussion

615.	A body of mass `0.5kg` is performing `SHM` with a time period `.^(pi)//_(2)` seconds. If its velocity at mean position is `1ms^(-1)`, then restoring force acting on the body at a phase angle `60^(@)` from exterme position isA. `0.5N`B. `1N`C. `2N`D. `4N`
Answer» Correct Answer - B `K = (4pi^(2)m)/(T^(2)), V_(max) = A omega, F=Kx = KA Cos .(pi)/(3)`

Discussion

616.	A body executing `SHM` has a maximum velocity of `1ms^(-1)` and a maximum acceleration of `4ms^(-1)`. Its time period of oscillation isA. `3.14s`B. `1.57s`C. `6.28s`D. `0.25s`
Answer» Correct Answer - B `omega = (a_(max))/(v_(max)) = (2T)/(T)`

Discussion

617.	Two clocks one working with the principle of oscillating pendulum, the other with that of oscillating spring are taken to the moon, thenA. both show correct time one the moonB. first one only shows correct timeC. second one only shows correct timeD. both show wrong time
Answer» Correct Answer - C

Discussion

618.	An oscillating mass spring system has mechanical energy `1` joule, when it has an amplitude `0.1m` and maximum speed of `1ms^(-1)`. The force constant of the spring is (in `Nm^(-1))`A. `100 N//m`B. `200 N//m`C. `300 N//m`D. `50 N//m`
Answer» Correct Answer - B `T.E_(max)=(1)/(2)KA^(2)` `1=(1)/(2)xxKxx0.01` `therefore K=(2)/(0.01)=200N//m`

Discussion

619.	An oscillating mass spring system has mechanical energy `1` joule, when it has an amplitude `0.1m` and maximum speed of `1ms^(-1)`. The force constant of the spring is (in `Nm^(-1))`A. `100`B. `200`C. `300`D. `50`
Answer» Correct Answer - B `KE_(max) = (1)/(2) KA^(2) (k = m omega^(2))`

Discussion

620.	Two pendulums oscillates with a constant phase difference of `90^(@)` and same amplitude. The maximum velocity of one pendulum is v, then the maximum velocity of the other pendulum will beA. 2 vB. vC. `v sqrt(2)`D. `sqrt(2v)`
Answer» Correct Answer - B The phase difference is constant and hence its period must be equal therefore velocities are also equal.

Discussion

621.	The time period of an oscillating spring of mass 630 g and spring constant 100 N/m with a load of 1 kg isA. `0.2pi s`B. `0.21pi s`C. `0.22pi s`D. `0.02pi s`
Answer» Correct Answer - C `T=2pisqrt((1+(0.63)/(3))/(100))=2pisqrt((1.21)/(100))=2pixx(1.1)/(10)` `=0.22pis`

Discussion

622.	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)/(sqrt(5))`D. `2T sqrt(5)`
Answer» Correct Answer - C `(T_(2))/(T_(1))=sqrt((g_(1))/(g_(2)))=sqrt((9)/(9+(9)/(4)))=sqrt((4)/(5))` `therefore T_(2)=(2)/(sqrt(5))T`.

Discussion

623.	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)/(sqrt(2))`C. `sqrt(2)T`D. `(T)/(2^(1//4))`
Answer» Correct Answer - D `(T_(2))/(T_(1))=sqrt((g)/(sqrt(g^(2)+g^(2))))=sqrt((g)/(gsqrt(2)))` `T_(2)=(T)/(sqrt(2))=(T)/(2^(1//4))`

Discussion

624.	Assertion: In damped oscillation, the energy of the system is dissipated continuously. Reason: For small damping, the oscillations remain approximately periodic.A. If both assertion and reson are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - B In damped oscillation like the motion of simple pendulum swinging in air, motion dies out eventually. This is because of the air drag and the friction at the support oppose the motion of the pendulukm and dissipate its energy gradually.

Discussion

625.	If a particle executes SHM of time period 4 s and amplitude 2 cm, find its maximum velocity and that at half its full displacement also find the acceleration at the turning points and when the displacement is 0.75 cm
Answer» Correct Answer - `3.14cms^(-1),2.72cms^(-1),4.93cm^(-2),1.85cms^(-2)`

Discussion

626.	A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 100N/m. the block is pulled to a distance x=10 cm from its equilibrium position at x=0 on a frictionless surface from rest at t=0. the kinetic energy and potential energy of the block when it is 5 cm away from the mean position isA. 0.0375 J, 0.125 JB. 0.125 J, 0.375 JC. 0.125 J, 0.125 JD. 0.375 J, 0.375 J
Answer» Correct Answer - A Here, `m=1kg,k=100" N "m^(-1)` A=10cm=0.1m The blocks executes SHM, its angular frequency is given by `omega=sqrt((k)/(m))=sqrt((100" N "m^(-1))/(1kg))=10" rad "s^(-1)` velocity of the block at x=5cm=0.05m is `v=omegasqrt(A^(2)-x^(2))=10sqrt((0.1)^(2)-(0.05)^(2))=10sqrt(7.5xx10^(-3))ms^(-1)` Kinetic energy of the block, `K=(1)/(2)mv^(2)=(1)/(2)xx1xx0.75=0.0375J` Potential energy of the block, `U=(1)/(2)kx^(2)=(1)/(x)xx100xx(0.05)^(2)=0.125J`

Discussion

627.	When a particle oscillates simple harmonically, its kinetic energy varies periodically. If frequency of the particle is n, the frequency of the kinetic energy isA. 4 nB. nC. 2 nD. `n//2`
Answer» Correct Answer - C `K.E,=(1)/(2) momega^(2) (A^(2)-x^(2))` `=(1)/(2) momega^(2) (A^(2)-A^(2) sin^(2)omegat)` `=(1)/(2) momega^(2)A^(2)(1-sin^(2)omega^(2)omegat)` `K.E.=(1)/(2)momega^(2)A^(2) ((1+cos2omegat))/(2)` `therefore` The frequency of its kinetic energy is `2omega`

Discussion

628.	If a particle executes an undamped S.H.M. of period T, then the period with which the total energy fluctuate isA. TB. 2TC. `T//2`D. `oo`
Answer» Correct Answer - D Totalo energy remains constant. It should not fluctuate `therefore T is oo`

Discussion

629.	If a particle executes an undamped S.H.M. of period of T, then the period with which the kinetic energy fluctuate isA. TB. 2TC. T/2D. `oo`
Answer» Correct Answer - C

Discussion

630.	A spring 60 cm long is stretched 2 cm, when subjected to a force of 20 gram weight what would be the length, when a force of 500 gram weight is applied.?
Answer» Correct Answer - 110cm

Discussion

631.	A particle executes simple harmonic motion between `x = -A and x = + A`. The time taken for it to go from `0 to A//2 is T_1 and to go from A//2 to (A) is (T_2)`. Then.A. `T_(1) lt T_(2)`B. `T_(1) gt T_(2)`C. `T_(1)=T_(2)`D. `T_(1)=2T_(2)`
Answer» Correct Answer - A Any SHM is given by the equation `x=sinomegat`, where x is the displacement of the body any instant t. A is the amplitude and `omega` is the angular frequency. when `x=0,omegat_(1)=0` `thereforet_(1)=0` when `x=A//2,omegat_(2)=pi//6,t_(2)=pi//6omega` when `x=A,omegat_(3)=pi//2,t_(3)=pi//2omega` Time taken from o to A/2 will be `t_(2)-t_(1)=(pi)/(6omega)=T_(1)` time taken from A/2 to A will be `t_(3)-t_(2)=(pi)/(2omega)-(pi)/(6omega)=(2pi)/(6omega)=(pi)/(3omega)=T_(2)` Hence `T_(2)gtT_(1)`

Discussion

632.	The periodic time of a linear harmonic oscillator is 2π seconds, with maximum displacement of 1 cm. If the particle starts from an extreme position, find the displacement of the particle after π/3 seconds.
Answer» Data : T = 2π s, A = 1 cm, t = π/3 ω = \(\frac{2\pi}T\) = \(\frac{2\pi}{2\pi}\) = 1 rad/s x = A cos ωt (∵ particle starts from extreme position) = (1) cos \((1\times\frac{\pi}3)\) = cos \((1\times\frac{\pi}3)\) = cos \(\frac{\pi}3\) = \(\frac{1}2\) cm

632.

The periodic time of a linear harmonic oscillator is 2π seconds, with maximum displacement of 1 cm. If the particle starts from an extreme position, find the displacement of the particle after π/3 seconds.

Answer»

Data : T = 2π s, A = 1 cm, t = π/3

ω = \(\frac{2\pi}T\) = \(\frac{2\pi}{2\pi}\) = 1 rad/s

x = A cos ωt (∵ particle starts from extreme position)

= (1) cos \((1\times\frac{\pi}3)\) = cos \((1\times\frac{\pi}3)\) = cos \(\frac{\pi}3\) = \(\frac{1}2\) cm

Discussion

633.	A particle executes SHM of period 12 seconds and amplitude 8 cm. Find time it takes to travel 3 cm from the positive extremity of its oscillation.
Answer» Correct Answer - 1.71s

Discussion

634.	A particle executes simple harmonic oscillation with an amplitudes a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position isA. `(T)/(4)`B. `(T)/(8)`C. `(T)/(12)`D. `(T)/(2)`
Answer» Correct Answer - C

Discussion

635.	The average acceleration of a particle performing SHM over one complete oscillation isA. `(omega^(2)A)/(2)`B. `(omega^(2)A)/(sqrt(2))`C. zeroD. `A omega^(2)`
Answer» Correct Answer - C The average acceleration of a particle performing SHM over one complete oscillation is zero.

Discussion

636.	Two simple harmonic motions of angular frequency `100 rad s^(-1)` and `1000 rad s^(-1)` have the same displacement amplitude. The ratio of their maximum accelerations isA. `1:10`B. `1:10^(2)`C. `1:10^(3)`D. `1:10^(4)`
Answer» Correct Answer - B

Discussion

637.	What is the time period for the function f(t) = sin ωt + cos ωt may represent the simple harmonic motion ?
Answer» Time period : T =2π/ω

637.

What is the time period for the function f(t) = sin ωt + cos ωt may represent the simple harmonic motion ?

Answer»

Time period :

T =2π/ω

Discussion

638.	Two simple harmonic motions of angular frequency `100 rad s^(-1)` and `1000 rad s^(-1)` have the same displacement amplitude. The ratio of their maximum accelerations isA. `1 : 10`B. `1 : 10^(2)`C. `1 : 10^(3)`D. `1 : 10^(4)`
Answer» Correct Answer - B Acceleration of simple harmonic motion is `a_(max)=-omega^(2)A` or `((a_(max))_(1))/((a_(max))_(2))=(omega_(1)^(2))/(omega_(2)^(2))" "("as A remains same")` or `((a_(max))_(1))/((a_(max))_(2))=((100)^(2))/((1000)^(2))=((1)/(10))^(2)=1:10^(2)`

Discussion

639.	A vibrating simple pendulum of period T is placed in a lift which is accelerating downwards. What will be the effect on the time period?
Answer» Time period increases as effective value of acceleration due to gravity decreases.

Discussion

640.	Consider two simple harmonic motion along x and y-axis having same frequencies but different amplitudes as x = A sin (ωt + φ) (along x axis) and y = B sin ωt (along y axis).Then show that \(\frac{x^2}{A^2}\) + \(\frac{y^2}{B^2}\) -\(\frac{2xy}{AB}\)cos ϕ = sin2 ϕ and also discuss the special casesNote: when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
Answer» (a) \(y = \frac{B}{A}x\) , equation is a straight line passing through origin with positive slope. (b) \(y = -\frac{B}{A}x\), equation is a straight line passing through origin with negative slope. (c) \(\frac{x^2}{A^2}\) + \(\frac{y^2}{B^2}\) = 1, equation is an ellipse whose center is origin. A2 B2 (d) \(x^2 + y^2 = A^2\) , equation is a circle whose center is origin. (e) \(\frac{x^2}{A^2}\) + \(\frac{y^2}{B^2}\) - \(\frac{2xy}{AB} \frac{1}{\sqrt{2}}\) = \(\frac{1}{2}\), equation is an ellipse which (oblique ellipse which means tilted ellipse)

640.

Consider two simple harmonic motion along x and y-axis having same frequencies but different amplitudes as x = A sin (ωt + φ) (along x axis) and y = B sin ωt (along y axis).Then show that \(\frac{x^2}{A^2}\) + \(\frac{y^2}{B^2}\) -\(\frac{2xy}{AB}\)cos ϕ = sin2 ϕ and also discuss the special casesNote: when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.

Answer»

(a) \(y = \frac{B}{A}x\) , equation is a straight line passing through origin with positive slope.

(b) \(y = -\frac{B}{A}x\), equation is a straight line passing through origin with negative slope.

(c) \(\frac{x^2}{A^2}\) + \(\frac{y^2}{B^2}\) = 1, equation is an ellipse whose center is origin. A2 B2

(d) \(x^2 + y^2 = A^2\) , equation is a circle whose center is origin.

(e) \(\frac{x^2}{A^2}\) + \(\frac{y^2}{B^2}\) - \(\frac{2xy}{AB} \frac{1}{\sqrt{2}}\) = \(\frac{1}{2}\), equation is an ellipse which (oblique ellipse which means tilted ellipse)

Discussion

641.	A particle executing simple harmonic motion has a kinetic energy K0 cos2 ωt. The maximum values of the potential energy and the total energy are, respectively ….(a) k0 /2 and k0 (b) k0 and 2k0 (c) k0 and k0 (d) 0 and 2 k0
Answer» Correct answer is (c) k₀ and k₀

Discussion

642.	Which one of the following equations of motion represents simple harmonic motion?(a) Acceleration = – k0 x + k1 x2 (b) Acceleration = – k(x + a)(c) Acceleration = k(x + a) (d) Acceleration = kx
Answer» (b) Acceleration = k(x + a)

Discussion

643.	The bob of vibrating simple pendulum is made of ice. How will the period of swing change when the ice starts melting?
Answer» The period of swing of simple pendulum will remain unchanged till the location of the centre of gravity of the bob left after melting the ice remains at a fixed distance from the point of suspension. If the centre of gravity of ice bob after melting is raised upwards, then the effective length of pendulum decreases and hence the time period of swing decrease. If the centre of gravity of ice shifts on lower side, the time period of swing increases.

643.

The bob of vibrating simple pendulum is made of ice. How will the period of swing change when the ice starts melting?

Answer»

The period of swing of simple pendulum will remain unchanged till the location of the centre of gravity of the bob left after melting the ice remains at a fixed distance from the point of suspension.

If the centre of gravity of ice bob after melting is raised upwards, then the effective length of pendulum decreases and hence the time period of swing decrease. If the centre of gravity of ice shifts on lower side, the time period of swing increases.

Discussion

644.	The function sin2 (ωt) represents (a) a simple harmonic motion with a period π/ω (b) a simple harmonic motion with a period 2π/ω (c) a periodic, but not simple harmonic motion with a period π/ω (d) a periodic, but not simple harmonic motion with a period 2π/ω
Answer» (a) a simple harmonic motion with a period π/ω sin²ωt = \(\frac{1}{2}\)(1 - cos 2ωt) This is an SHM with period \(\frac{2π}{2ω}= \frac{π}{ω}\). Its equilibrium position is at \(\frac{1}{2}\) Instead of zero.

644.

The function sin2 (ωt) represents (a) a simple harmonic motion with a period π/ω (b) a simple harmonic motion with a period 2π/ω (c) a periodic, but not simple harmonic motion with a period π/ω (d) a periodic, but not simple harmonic motion with a period 2π/ω

Answer»

(a) a simple harmonic motion with a period π/ω

sin²ωt = \(\frac{1}{2}\)(1 - cos 2ωt)

This is an SHM with period \(\frac{2π}{2ω}= \frac{π}{ω}\).

Its equilibrium position is at \(\frac{1}{2}\) Instead of zero.

Discussion

645.	The bob of a vibrating simple pendulum is made of ice. How will the time period change when the ice starts melting?
Answer» The time period is independent of the mass of bob.

Discussion

646.	STATEMENT -1 `:` The time period of osciallation of a simple pendulum of constant length is more at a place inside a mine than on the surface of the earth. and STATEMENT -2 `:` The frequency of osciallations ofa simple pendulum is more at aplace inside a mine than on the surface of the earth .
Answer» Correct Answer - 3

Discussion

647.	A pendulum clock is taken to the bottom of a deep mine. Will it gain or lose time? How should its length be altered to correct the time?A. looses time, length to be increasedB. looses time, length to be decreasedC. gains time, length to be increasedD. gains time, length to be decreased
Answer» Correct Answer - B

Discussion

648.	The ratio of the height of the mountain and the depth of a mine, if a pendulum swings with the same period at the top of the mountain and at the bottom of the mine isA. `1 : 1`B. `1 : 2`C. `2 : 1`D. `4 : 1`
Answer» Correct Answer - B `T_(n)=T_(d)` `2pisqrt((l)/(g_(h)))=2pisqrt((l)/(g_(d)))` `g_(h)=g_(d)` `1-(2h)/(R)=1-(d)/(R)` `2h=d` `(h)/(d)=(1)/(2)`

Discussion

649.	Simple harmonic motion is represented by, y = a sin (ωt + θ). If θ = π/6, then for what value of y, the body starts oscillating?
Answer» θ = π/6 and t = 0, when the body starts oscillating. Hence, y = a sin π/6 = a/2.

Discussion

650.	When a pendulum clock gains time, what adjustment should be made?
Answer» When a pendulum clock gains time (goes fast), it makes more vibrations than required per day. That means, its period of oscillation has decreased. So, in order to correct it, the length of the pendulum should be properly increased.

Discussion

Explore topic-wise InterviewSolutions in .

A body executes `SHM` has its velocity `10cm//s` and `7cm//s` when its displacements from mean positions are `3cm` and `4cm` respectively. The length of path is nearlyA. `10cm`B. `9.5cm`C. `4cm`D. `11.36cm`

In a simple pendulum if iron sphere is replaced by a wooden sphere of same mass its time period of oscillation willA. increasesB. decreasesC. remains sameD. It does not oscillate

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

A particle performing `SHM` with a frequency of `5Hz` and amplitude `2cm` is initially at position exterme position. The equation for its displacement (in metre) isA. `x = 0.02 sin 10 pit`B. `x = 0.02 sin 5 pit`C. `x = 0.02 cos 10 pit`D. `x = 0.02 cos 5pi t`

A particle executing `SHM` from exterme position towards centre is observed to be at distances `x_(1),x_(2)` and `x_(3)` from the centre at the end of three successive seconds. The period of `SHM` is.

`A` mass `M` is suspended from a light spring. An additional mass `m` added to it displaces the spring further by distance `x` then its time period isA. `T = 2pi sqrt((mg)/(x(M+m)))`B. `T = 2pi sqrt(((M+m)x)/(mg))`C. `T = 2pi sqrt(((M+m)x)/(Mg))`D. `T = 2pi sqrt((Mg)/(mg))`

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

A body executes `SHM`, such that its velocity at the mean position is `1ms^(-1)` and acceleration at exterme position is `1.57 ms^(-2)`. Calculate the amplitude and the time period of oscillation.

A body of mass `0.5kg` is performing `SHM` with a time period `.^(pi)//_(2)` seconds. If its velocity at mean position is `1ms^(-1)`, then restoring force acting on the body at a phase angle `60^(@)` from exterme position isA. `0.5N`B. `1N`C. `2N`D. `4N`

A body executing `SHM` has a maximum velocity of `1ms^(-1)` and a maximum acceleration of `4ms^(-1)`. Its time period of oscillation isA. `3.14s`B. `1.57s`C. `6.28s`D. `0.25s`

Two clocks one working with the principle of oscillating pendulum, the other with that of oscillating spring are taken to the moon, thenA. both show correct time one the moonB. first one only shows correct timeC. second one only shows correct timeD. both show wrong time

An oscillating mass spring system has mechanical energy `1` joule, when it has an amplitude `0.1m` and maximum speed of `1ms^(-1)`. The force constant of the spring is (in `Nm^(-1))`A. `100 N//m`B. `200 N//m`C. `300 N//m`D. `50 N//m`

An oscillating mass spring system has mechanical energy `1` joule, when it has an amplitude `0.1m` and maximum speed of `1ms^(-1)`. The force constant of the spring is (in `Nm^(-1))`A. `100`B. `200`C. `300`D. `50`

Two pendulums oscillates with a constant phase difference of `90^(@)` and same amplitude. The maximum velocity of one pendulum is v, then the maximum velocity of the other pendulum will beA. 2 vB. vC. `v sqrt(2)`D. `sqrt(2v)`

The time period of an oscillating spring of mass 630 g and spring constant 100 N/m with a load of 1 kg isA. `0.2pi s`B. `0.21pi s`C. `0.22pi s`D. `0.02pi s`

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)/(sqrt(5))`D. `2T sqrt(5)`

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)/(sqrt(2))`C. `sqrt(2)T`D. `(T)/(2^(1//4))`

If a particle executes SHM of time period 4 s and amplitude 2 cm, find its maximum velocity and that at half its full displacement also find the acceleration at the turning points and when the displacement is 0.75 cm

When a particle oscillates simple harmonically, its kinetic energy varies periodically. If frequency of the particle is n, the frequency of the kinetic energy isA. 4 nB. nC. 2 nD. `n//2`

If a particle executes an undamped S.H.M. of period T, then the period with which the total energy fluctuate isA. TB. 2TC. `T//2`D. `oo`

If a particle executes an undamped S.H.M. of period of T, then the period with which the kinetic energy fluctuate isA. TB. 2TC. T/2D. `oo`

A spring 60 cm long is stretched 2 cm, when subjected to a force of 20 gram weight what would be the length, when a force of 500 gram weight is applied.?

A particle executes simple harmonic motion between `x = -A and x = + A`. The time taken for it to go from `0 to A//2 is T_1 and to go from A//2 to (A) is (T_2)`. Then.A. `T_(1) lt T_(2)`B. `T_(1) gt T_(2)`C. `T_(1)=T_(2)`D. `T_(1)=2T_(2)`

The periodic time of a linear harmonic oscillator is 2π seconds, with maximum displacement of 1 cm. If the particle starts from an extreme position, find the displacement of the particle after π/3 seconds.

A particle executes SHM of period 12 seconds and amplitude 8 cm. Find time it takes to travel 3 cm from the positive extremity of its oscillation.

A particle executes simple harmonic oscillation with an amplitudes a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position isA. `(T)/(4)`B. `(T)/(8)`C. `(T)/(12)`D. `(T)/(2)`

The average acceleration of a particle performing SHM over one complete oscillation isA. `(omega^(2)A)/(2)`B. `(omega^(2)A)/(sqrt(2))`C. zeroD. `A omega^(2)`

Two simple harmonic motions of angular frequency `100 rad s^(-1)` and `1000 rad s^(-1)` have the same displacement amplitude. The ratio of their maximum accelerations isA. `1:10`B. `1:10^(2)`C. `1:10^(3)`D. `1:10^(4)`

What is the time period for the function f(t) = sin ωt + cos ωt may represent the simple harmonic motion ?

Two simple harmonic motions of angular frequency `100 rad s^(-1)` and `1000 rad s^(-1)` have the same displacement amplitude. The ratio of their maximum accelerations isA. `1 : 10`B. `1 : 10^(2)`C. `1 : 10^(3)`D. `1 : 10^(4)`

A vibrating simple pendulum of period T is placed in a lift which is accelerating downwards. What will be the effect on the time period?

A particle executing simple harmonic motion has a kinetic energy K0 cos2 ωt. The maximum values of the potential energy and the total energy are, respectively ….(a) k0 /2 and k0 (b) k0 and 2k0 (c) k0 and k0 (d) 0 and 2 k0

Which one of the following equations of motion represents simple harmonic motion?(a) Acceleration = – k0 x + k1 x2 (b) Acceleration = – k(x + a)(c) Acceleration = k(x + a) (d) Acceleration = kx

The bob of vibrating simple pendulum is made of ice. How will the period of swing change when the ice starts melting?

The function sin2 (ωt) represents (a) a simple harmonic motion with a period π/ω (b) a simple harmonic motion with a period 2π/ω (c) a periodic, but not simple harmonic motion with a period π/ω (d) a periodic, but not simple harmonic motion with a period 2π/ω

The bob of a vibrating simple pendulum is made of ice. How will the time period change when the ice starts melting?

STATEMENT -1 `:` The time period of osciallation of a simple pendulum of constant length is more at a place inside a mine than on the surface of the earth. and STATEMENT -2 `:` The frequency of osciallations ofa simple pendulum is more at aplace inside a mine than on the surface of the earth .

A pendulum clock is taken to the bottom of a deep mine. Will it gain or lose time? How should its length be altered to correct the time?A. looses time, length to be increasedB. looses time, length to be decreasedC. gains time, length to be increasedD. gains time, length to be decreased

The ratio of the height of the mountain and the depth of a mine, if a pendulum swings with the same period at the top of the mountain and at the bottom of the mine isA. `1 : 1`B. `1 : 2`C. `2 : 1`D. `4 : 1`

Simple harmonic motion is represented by, y = a sin (ωt + θ). If θ = π/6, then for what value of y, the body starts oscillating?

When a pendulum clock gains time, what adjustment should be made?