956 + Interview Questions in OSCILLATIONS IN GENERAL AWARENESS Page 5 InterviewSolution

201.	A particle which is attached to a spring oscillates horizontally with simple harmonic motion with a frequency of `1//pi` Hz and total energy of 10J. If the maximum speed of the particle is `0.4ms^(-1)`, what is the force constant of the spring? What will be the maximum potential energy of the spring during the motion?
Answer» Correct Answer - `k=500Nm^(-1),U_(max)=10J`

Discussion

202.	A body oscillates with a simple harmonic motion having amplitude 0.05 m. At a certain instant of time, its displacement is 0.01 m and acceleration is `1.0 m//s^(2)`. The period of oscillation isA. 0.1 sB. 0.2 sC. `pi//10s`D. `pi//5s`
Answer» Correct Answer - D Acceleration `=omega^(2)x` `1=omega^(2)xx0.01` `therefore omega^(2)=100 therefore omega=10" but "T=(2pi)/(omega)` `therefore T=(pi)/(5)s`

Discussion

203.	The period of a simple harmonic oscillator is 2 s. If it crosses mean position at an instant. The time after which its displacement from the mean position will be half of the amplitude is.A. `(1//8)s`B. `(1//6)s`C. `(1//4)s`D. `(1//2)s`
Answer» Correct Answer - B `x=Asinomegat` `(1)/(2)A=Asinomegat` `omegat=sin^(-1)((1)/(2))=(pi)/(6)` `(2pi)/(T)t=(pi)/(6)` `t=(T)/(12)=(2)/(12)=(1)/(6)s`

Discussion

204.	Two astronauts on the surface of moon cannot talk to each other. Why?
Answer» This is because moon has no atmosphere and sound cannot travel in vacuum.

Discussion

205.	A vertical pole of length `l`, density `rho`, area of cross section `A`, floats in two immiscible liquids of densities `rho_(1)` and `rho_(2)`. In equilibrium position the bottom end is at the interface of the liquids. When the cylinder is displaced vertically, the time period of oscillation is.......... A. `pisqrt((rhol)/((rho_(1)+rho_(2))g))`B. `pi sqrt((rhol)/(g)) ((1)/(sqrt(rho_(1)))+(1)/(sqrt(rho_(2))))`C. `pi sqrt(((rho_(1)+rho_(2))l)/(rhog))`D. `pi sqrt((l)/(rhog)) (sqrt(rho_(1))+sqrt(rho_(2)))`
Answer» Correct Answer - B Here interface acts as mean position of `SHM`: `:.` time period, `T = (T_(1))/(2) +(T_(2))/(2)` `T_(1) = 2pi sqrt((rhol)/(rho_(2)g)): T_(2) = 2pi sqrt((rhol)/(rho_(1)g))`

Discussion

206.	A block of mass 2 kg executes SHM on a smooth horizontal plane under the action of restoring force of a spring. If the amplitude and time period of oscillation are 0.05 m and `pi//5` second, the maximum force exerted by the spring on the block isA. 5 NB. 10 NC. 100 ND. 19.6 N
Answer» Correct Answer - B `omega=(2pi)/(T)=(2pi)/(pi//5)` `omega=10" rad/s"` `F=momega^(2)A` `=2xx10^(2)xx0.05` `=10N`

Discussion

207.	If the differential equation given by `(d^(2)y)/(dt^(2))+2k(dy)/(dt)+omega^(2)y=F_(0)sin pi t` describes the oscillatory motion of body in a dissipative medium under the influence of a periodic force, then the state of maximum amplitude of the oscillation is a measure ofA. free vibrationB. damped vibrationC. forced vibrationD. resonance
Answer» Correct Answer - D Here, the state of maximum amplitude of the oscillation is a measure of resonance.

Discussion

208.	Maximum kinetic energy of a particle of mass 1 kg in SHM is 8 J. Time period of SHM is 4 s. Maximum potential energy during the motion is 10 J. ThenA. amplitude of oscillations is approximately 3.53 mB. minimum potential energy of the particle is 4 JC. maximum acceleration of the particle is approximately 6.3 `ms^(-2)`D. minimum kinetic energy of the particle is 2 J
Answer» Correct Answer - C Maximum kinetic energy = energy of oscillation in SHM `therefore" "8=(1)/(2)kA^(2)` `therefore" "kA^(2)=16" "...(i)` Further, `2pi sqrt((m)/(k))=4 rArr K = (pi^(2))/(4)" "...(ii)` From Eqs. (i) and (ii), we get `k = 2.4 Nm^(-1)and A = 2.53m` Maximum acceleration of the particle will be `a_(max)=omega^(2)A=(k)/(m)A=((25)/(1))(2.53)=6.3 ms^(-2)`

Discussion

209.	Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time their displacement is one third their amplitude. What is the phase difference between them?
Answer» Correct Answer - B

Discussion

210.	Two particles execute simple harmonic motion of same amplitude and frequency along the same straight line. They pass on another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?A. `30^(@)`B. `60^(@)`C. `90^(@)`D. `120^(@)`
Answer» Correct Answer - D Let two simple harmonic motions are y = a sin `omega`t and y = a sin `(omega t + phi)`. In the first case, `(a)/(2)=a sin omega t` `rArr" "sin omega t = (1)/(2)` `therefore" "cos omega t=(sqrt(3))/(2)` In the second case, `(a)/(2)=a sin (omega t + phi)` `rArr" "(1)/(2)=[sin omegat.cos phi+cos omega t sin phi]` `rArr" "(1)/(2)=[(1)/(2)cos phi + (sqrt(3))/(2)sin phi]` `rArr" "1 - cos phi = sqrt(3) sin phi` `rArr" "(1-cos phi)^(2)=3 sin^(2)phi` `rArr" "(1-cos phi)^(2)=3(1-cos^(2)phi)` On solving above equation, we get `cos phi = + 1` or `cos phi = (-1)/(2)` i.e. `phi = 0` or `phi = 120^(@)`

Discussion

211.	Why does the sound travel faster in humid air ?
Answer» Because the density of water vapour is less than that of the dry air hence density of air decreases with the increase of water vapours or humidity and velocity of sound inversely proportional to square root of density.

Discussion

212.	In which medium do the sound waves travel faster-solids, liquids or gases? Why?
Answer» Sound travels in solids with travel highest velocity. This is because coefficient of elasticity of solids is much greater than coefficient of elasticity of liquids and gases.

Discussion

213.	Assertion: An earthquake will not cause uniform damage to all building in an affected area, even if they are built with the same strength and materials. Reason: The one with its natural frequency close to the frequency of seismic wave is likely to be damaged less.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 - C An earthquake will not cause uniform damage to all building in an affected area because the natural frequencies of abuilding depend on its height and other size parameters and the nature of building materials. The one with its natural frequency close to the frequency of seismic wave is likely to be damaged more.

Discussion

214.	A body of mass 0.1 kg is executing SHM of amplitude 1.0 m and period 0.2 s. what is 2 cm. find the time period.
Answer» Correct Answer - 98.7 N

Discussion

215.	A particle executing SHM has an acceleration of `0.5cms^(-1)` when its displacement is 2 cm. find the time period.
Answer» Correct Answer - `4pis`

Discussion

216.	A pendulum has a string of length 99.39 cm. how much length of the pendulum must be shortened to keep the current time of the pendulum if it loses 4 s a day?A. 0.0009 cmB. 0.009 cmC. 0.09 cmD. 0.9 cm
Answer» Correct Answer - B Time lost by pendulum per day=4s, dn=-4s number of seconds in a day, n=`24xx3600s` Let dl be the length by which the pendulum should be shortened to keep the correct time `(dn)/(n)=-(dl)/(2l)implies(-4)/(24xx3600)=(-dl)/(2xx99.39)` `impliesdl=0.00920cm`

Discussion

217.	A simple harmonic motion is represented by `x = 12 sin (10 t + 0.6)` Find out the maximum acceleration, if displacement is measured in metres and time in seconds.A. `-1200 m`B. `-2000 m`C. `-2400 m`D. `7200 m`
Answer» Correct Answer - A Given equation, `x = 12 sin (10 t + 0.6)` On comparing with `x(t)=A sin (omega t+phi)` We have, Amplitude, A = 12 m Angular frequency, `omega = 10 rads^(-1)` `therefore` Maximum acceleration, `a_(max)=-omega^(2)A` `=-(10)^(2)xx12` `= -1200ms^(-2)`

Discussion

218.	At what position the KE of an oscillating simple pendulum Is maximum?
Answer» Kinetic energy is maximum at the mean position.

Discussion

219.	Mention the relation between period and frequency of periodic.
Answer» Frequency f = 1/T.

Discussion

220.	The velocity of the particle midway between mean and extreme position, performing S.H.M. isA. `A omega`B. `(sqrt(3))/(2)A omega`C. `(2)/(sqrt(2))A omega`D. `sqrt(3)A omega`
Answer» Correct Answer - B `v_(m)=omegasqrt(A^(2)-x^(2))=omegasqrt(A^(2)-(A^(2))/(4))` `=omegasqrt((4A^(2)-A^(2))/(4)` `=omegasqrt((3A^(2))/(4))=sqrt(3)/(2)Aomega`

Discussion

221.	A particle executing SHM has a velocity of 2 `ms^(-1)` when its displacement from mean position is 1 cm and a velocity of `1 ms^(-1)` when its displacement is 2 cm. Its amplitude of oscillationisA. 5 cmB. `sqrt(5) cm`C. 3 cmD. `sqrt(7) cm`
Answer» Correct Answer - B `(V_(1))/(V_(2))=sqrt((A^(2)-y_(1)^(2))/(A^(2)-y_(2)^(2))),(2)/(1)=sqrt((A^(2)-1)/(A^(2)-2^(2)))` `4A^(2)-16=A^(2)-1` `3A^(2)=15,A^(2)=5` `A=sqrt((2^(2)xx2^(2)-1^(2)xx1^(2))/(2^(2)-1))` `=sqrt((15)/(3))=sqrt(5)` By shortcut method, `A=sqrt(v_(1)^(2)+v_(2)^(2))`

Discussion

222.	If a particle performing S.H.M. with a period of T s, then the time required to reach from midway to extreme position will beA. `(T)/(2)s`B. `(T)/(4)s`C. `(T)/(6)s`D. `(T)/(12)s`
Answer» Correct Answer - C `t=(T)/(12)`

Discussion

223.	The period of a body performing S.H.M. of frequency 5 Hz isA. `(2pi//5) s`B. `(5pi) s`C. `(pi//5) s`D. `(0.2) s`
Answer» Correct Answer - D `T=(1)/(n)=(1)/(5)=0.2s`.

Discussion

224.	The average displacement over a period of SHM is (A = amplitude of SHM) (A) 0 (B) A (C) 2A(D) 4A.
Answer» Correct option is (A) 0

Discussion

225.	The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are 2 m/s and` 4 m//s^(2)`. Then angular velocity will beA. 3 `rads^(-1)`B. 0.5 `rads^(-1)`C. 1 `rads^(-1)`D. 2 `rads^(-1)`
Answer» Correct Answer - D At extreme position, displacement of particle executing SHM is maximum. Maximum velocity of particle executing SHM is given by `therefore" "v_(max) = omega A` Maximum acceleration of particle executing SHM is given by, `a =-omega^(2)A or a_(max)=\|A\|=omega^(2)A` As `a_(max)=(Aomega)omega=v_(max).omega` `therefore" "omega=(a_(max))/(v_(max))=(4)/(2)=2 rads^(-1)`

Discussion

226.	The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are 2 m/s and` 4 m//s^(2)`. Then angular velocity will beA. 1 rad/sB. 2 rad/sC. 4 rad/sD. 5 rad/s
Answer» Correct Answer - B `omega=(a_(m))/(v_(m))=(4)/(2)=2`

Discussion

227.	The total energy of a particle executing S.H.M. is proportional toA. square of amplitude of motionB. frequency of oscillationC. velocity in equilibrium positionD. displacement from equilibrium position
Answer» Correct Answer - A Total energy of a particle executing SHM is `E = (1)/(2)m omega a^(2) or E prop a^(2)` It means total energy is directly proportional to square of amplitude.

Discussion

228.	The average displacement over a period of S.H.M. is (A = amplitude of S.H.M)
Answer» Correct Answer - A

Discussion

229.	The potential energy of a particle with displacement X is `U(X)`. The motion is simple harmonic, when (K is a positive constant)A. `u=(-kx^(2))/(2)`B. `u=(1)/(2)kx^(2)`C. `u=k`D. `u=kx`
Answer» Correct Answer - B `P.E.=(1)/(2) m omega^(2)x^(2) " As" (1)/(2) m omega^(2)=`const. `therefore P.E. prop x^(2)=k x^(2)`

Discussion

230.	The phase difference between displacement and acceleration of a particle performing S.H.M. isA. `(pi)/(2)`B. `pi` radC. `2pi` radD. `(3pi)/(2)` rad
Answer» Correct Answer - B

Discussion

231.	A particle executing linear S.H.M. has velocities `v_(1)` and `v_(2)` at distances `x_(1)` and `x_(2)` respectively from the mean position. The angular velocity of the particle isA. `sqrt((x_(1)^(2)-x_(2)^(2))/(v_(2)^(2)-v_(1)^(2)))`B. `sqrt((v_(2)^(2)-v_(1)^(2))/(x_(1)^(2)-x_(2)^(2)))`C. `sqrt((x_(1)^(2)+x_(2)^(2))/(v_(2)^(2)+v^(2)))`D. `sqrt((v_(2)^(2)+v_(1)^(2))/(x_(2)^(2)+x_(1)^(2)))`
Answer» Correct Answer - B

Discussion

232.	Why is the term angular frequency (ω) used here for a linear motion ?
Answer» A linear SHM is the projection of a UCM on a diameter of the circle. The angular speed co of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.

Discussion

233.	The total energy of a particle executing S.H.M. is proportional toA. square of amplitudeB. square root of angular velocityC. amplitudeD. angular velocity
Answer» Correct Answer - A

Discussion

234.	State at which point during an oscillation the oscillator has zero velocity but positive acceleration ?
Answer» At the left extreme, i.e., x = – A, so that a = – ω² x = – ω²(- A) = ω²A = a_max

Discussion

235.	The period of a simple pendulum is doubled, whenA. its length is doubledB. the mass of the bob is doubledC. its length is four timesD. the amplitude of the pendulum is double and the length of the pendulum also double
Answer» Correct Answer - C

Discussion

236.	What is the effect on the time period of a simple pendulum if the mass of **** is doubled?A) HalvedB) DoubledC) Becomes eight timeD) No effect
Answer» D) No effect \(T=2\pi \sqrt \frac{L}{g}\) It can be seen that T is independent of mass.

236.

What is the effect on the time period of a simple pendulum if the mass of **** is doubled?A) HalvedB) DoubledC) Becomes eight timeD) No effect

Answer»

D) No effect

\(T=2\pi \sqrt \frac{L}{g}\)

It can be seen that T is independent of mass.

Discussion

237.	What is the nature of thermal changes in air, when a sound wave propagates through it?
Answer» When a sound wave travels through air, the changes in pressure and volume are adiabatic, i.e., temperature falls in the region of rarefaction.

Discussion

238.	Statement-I : Time period of spring block system is the same whether in an accelerated or in an inertial frame of reference. Statement-II : Mass of the block of spring block system and spring constant of the spring are independent of the acceleration of the frame of reference.A. Statement-I is true, Statement-Ii is true, Statement-II is a correct explanation for Statement-I.B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanantion for Statement-I.C. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true
Answer» Correct Answer - A

Discussion

239.	A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance `(2A)/3` from equilibrium position. The new amplitude of the motion is:A. `(A)/(3)sqrt(41)`B. `3A`C. `Asqrt(3)`D. `(7A)/(3)`
Answer» Correct Answer - D Applying conservation of energy `(1)/(2) m omega^(2) ((2A)/(3))^(2) + 9 xx (1)/(2) m omega^(2) [A^(2) -((2A)/(3))^(2)]` `= (1)/(2) m omega^(2) A_(1)^(2)` `rArr A_(1) = (7A)/(3)`

Discussion

240.	A particle moves in a circular path with a continuously increasing speed. Its motion isA. nonperiodicB. OscillatoryC. Simple harmonicD. UCM
Answer» Correct Answer - B

Discussion

241.	What are the Equation of beats ?
Answer» Equation of beats : If two waves of equal amplitudes ‘a’ and slightly different frequencies n₁ and n₂ travelling in a medium in the same direction then equation of beats is given by y = A sin π (n₁ – n₂)t where A = 2a cos π (n₁ – n₂)t = Amplitude of resultant wave. Amplitude of resultant wave. Beat frequency : n = n₁ – n₂. Beat period: 1/Beat frequency = 1/n₁ – n₂

241.

What are the Equation of beats ?

Answer»

Equation of beats : If two waves of equal amplitudes ‘a’ and slightly different frequencies n₁ and n₂ travelling in a medium in the same direction then equation of beats is given by

y = A sin π (n₁ – n₂)t where A = 2a cos π (n₁ – n₂)t = Amplitude of resultant wave. Amplitude of resultant wave.

Beat frequency : n = n₁ – n₂.

Beat period: 1/Beat frequency = 1/n₁ – n₂

Discussion

242.	You have a light spring, a metre scale and a known mass. How will you find time period of vibration of mass without the use of clock?
Answer» Attach the given mass (m) to the spring and suspend the spring. Measure the increase in length (l) of the spring with the help of the scale. Restoring force of the spring is given by, F = −kl = −mg Or \(\frac{m}{j}=\frac{l}{g}\) We know that time period of a vibrating system is given by, T = \(2\pi\sqrt{\frac{m}{k}}\) Or T = \(2\pi\sqrt{\frac{l}{g}}\)

242.

You have a light spring, a metre scale and a known mass. How will you find time period of vibration of mass without the use of clock?

Answer»

Attach the given mass (m) to the spring and suspend the spring. Measure the increase in length (l) of the spring with the help of the scale. Restoring force of the spring is given by,

F = −kl = −mg

Or \(\frac{m}{j}=\frac{l}{g}\)

We know that time period of a vibrating system is given by,

T = \(2\pi\sqrt{\frac{m}{k}}\)

Or T = \(2\pi\sqrt{\frac{l}{g}}\)

Discussion

243.	A particle is performing linear S.H.M. at a point A, on the path, its potential energy is three times kinetic energy. At another point B on the same path, its kinetic energy is 3 times the potential energy. The ratio of the potential energy at A to its potential energy at B isA. `9 : 1`B. `1 : 9`C. `1 : 3`D. `3 : 1`
Answer» Correct Answer - D `T.E._(A)=K.E_(A)+P.E_(A)` `=KE_(A)+3KE_(A)` `=4KE_(A)` `T.E_(B)=KE_(B)+KE_(B)=KE_(B)+(1)/(3)KE_(B)` `T.E_(B)=(4KE_(B))/(3)`

Discussion

244.	A body of mass 25 gm performs linear S.H.M. The force constant of the motion is 400 dyne/cm. When the body is at a distance of 10 cm from the equilibrium position has velocity of 40 cm/s, then the total energy of the body will beA. `40xx10^(4)J`B. `4xx10^(4)erg`C. `2xx10^(4)erg`D. `2xx10^(5)J`
Answer» Correct Answer - B `omega=sqrt((k)/(m))=sqrt((400)/(25))=(20)/(5)=4` `v=omegasqrt(A^(2)-x^(2))therefore 40=4sqrt(A^(2)-100)` `100=A^(2)-100 therefore A^(2)=200` `T.E.=(1)/(2)kA^(2)=(1)/(2)xx400xx200=4xx10^(4)erg`

Discussion

245.	When a force F1 acts on a particle, frequency is 6 Hz, when a force F2 acts, frequency is 8 Hz. What is the frequency when both the forces act simultaneously in the same direction?
Answer» Let m be the mass of the particle and r be its amplitude of oscillations. Then F₁ = mrω₁² = mr4π² v² and F₂ = mr4π² v₂² When both the forces act simultaneously, on the particle, then F = F₁ + F₂ or, mr4π² v² = mr4π² v₁² + mr4π²v₂² or, v² = v₁² + v₂² or, v = √{v₁² + v₂²} = √{6² + 8²} = 10 Hz

245.

When a force F1 acts on a particle, frequency is 6 Hz, when a force F2 acts, frequency is 8 Hz. What is the frequency when both the forces act simultaneously in the same direction?

Answer»

Let m be the mass of the particle and r be its amplitude of oscillations. Then

F₁ = mrω₁² = mr4π² v²

and F₂ = mr4π² v₂²

When both the forces act simultaneously, on the particle, then

F = F₁ + F₂

or, mr4π² v² = mr4π² v₁² + mr4π²v₂²

or, v² = v₁² + v₂²

or, v = √{v₁² + v₂²}

= √{6² + 8²} = 10 Hz

Discussion

246.	A particle is moving in a circle with uniform speed its motion isA. periodic and simple harmonicB. a periodicC. periodic but not simple harmonicD. non periodic but simple harmonic
Answer» Correct Answer - C

Discussion

247.	In simple harmonic motion, the quantities are not constantA. amplitude and frequencyB. potential energy and kinetic energyC. total energy and propagation constantD. path length
Answer» Correct Answer - B

Discussion

248.	The necessary and sufficient condition for S.H.M. isA. constant period and inertial propertyB. constant acceleration and elasticity propertyC. proportionality between equilibrium position in opposite directionD. periodic and harmonic
Answer» Correct Answer - C

Discussion

249.	The maximum velocity of a point undergoing simultaneous two oscillations given by `x_(1) =a cos wt` and `x_(2) = a cos 2 wt`is:A. `2.74 aw`B. `2aw`C. `1.414aw`D. `aw`
Answer» Correct Answer - A `x = a cos wt +a cos 2 wt` `v = (dx)/(dt)` For `v_(max), (dv)/(dt) = 0` `rArr cos wt = 0.64 :. V_(max) = 2.74 aw`

Discussion

250.	A particle executing SHM has a maximum speed of `30 cm//s` and a maximum acceleration of `60 cm/s^(2)` . The period of oscillation isA. `pisec`B. `(pi)/(2)sec`C. `2pisec`D. `(pi)/(t)sec `
Answer» Let equation of an SHM is represented by `y=asinomegat` `v=(dy)/(dt)=aomegacosomegat` `rArr(v)max=aomega=30`" " .....(i)` Acceleration `(A)=(dx^(2))/(dt^(2))=-aomega^(2)sinomegat` `A_(max)=omega^(2)a=60" " .....(ii)` Eqs. (i) and (ii), we get `omega(omegaa)=60rArromega(30)=60` `rArromega=2rad//s` `rArr(2pi)/(T)=2red//srArrT=pisec`

Discussion

Explore topic-wise InterviewSolutions in .

A body oscillates with a simple harmonic motion having amplitude 0.05 m. At a certain instant of time, its displacement is 0.01 m and acceleration is `1.0 m//s^(2)`. The period of oscillation isA. 0.1 sB. 0.2 sC. `pi//10s`D. `pi//5s`

The period of a simple harmonic oscillator is 2 s. If it crosses mean position at an instant. The time after which its displacement from the mean position will be half of the amplitude is.A. `(1//8)s`B. `(1//6)s`C. `(1//4)s`D. `(1//2)s`

Two astronauts on the surface of moon cannot talk to each other. Why?

A block of mass 2 kg executes SHM on a smooth horizontal plane under the action of restoring force of a spring. If the amplitude and time period of oscillation are 0.05 m and `pi//5` second, the maximum force exerted by the spring on the block isA. 5 NB. 10 NC. 100 ND. 19.6 N

Two particles execute SHM of same amplitude and frequency on parallel lines. They pass one another when moving in opposite directions each time their displacement is one third their amplitude. What is the phase difference between them?

Why does the sound travel faster in humid air ?

In which medium do the sound waves travel faster-solids, liquids or gases? Why?

A body of mass 0.1 kg is executing SHM of amplitude 1.0 m and period 0.2 s. what is 2 cm. find the time period.

A particle executing SHM has an acceleration of `0.5cms^(-1)` when its displacement is 2 cm. find the time period.

A pendulum has a string of length 99.39 cm. how much length of the pendulum must be shortened to keep the current time of the pendulum if it loses 4 s a day?A. 0.0009 cmB. 0.009 cmC. 0.09 cmD. 0.9 cm

A simple harmonic motion is represented by `x = 12 sin (10 t + 0.6)` Find out the maximum acceleration, if displacement is measured in metres and time in seconds.A. `-1200 m`B. `-2000 m`C. `-2400 m`D. `7200 m`

At what position the KE of an oscillating simple pendulum Is maximum?

Mention the relation between period and frequency of periodic.

The velocity of the particle midway between mean and extreme position, performing S.H.M. isA. `A omega`B. `(sqrt(3))/(2)A omega`C. `(2)/(sqrt(2))A omega`D. `sqrt(3)A omega`

A particle executing SHM has a velocity of 2 `ms^(-1)` when its displacement from mean position is 1 cm and a velocity of `1 ms^(-1)` when its displacement is 2 cm. Its amplitude of oscillationisA. 5 cmB. `sqrt(5) cm`C. 3 cmD. `sqrt(7) cm`

If a particle performing S.H.M. with a period of T s, then the time required to reach from midway to extreme position will beA. `(T)/(2)s`B. `(T)/(4)s`C. `(T)/(6)s`D. `(T)/(12)s`

The period of a body performing S.H.M. of frequency 5 Hz isA. `(2pi//5) s`B. `(5pi) s`C. `(pi//5) s`D. `(0.2) s`

The average displacement over a period of SHM is (A = amplitude of SHM) (A) 0 (B) A (C) 2A(D) 4A.

The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are 2 m/s and` 4 m//s^(2)`. Then angular velocity will beA. 3 `rads^(-1)`B. 0.5 `rads^(-1)`C. 1 `rads^(-1)`D. 2 `rads^(-1)`

The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are 2 m/s and` 4 m//s^(2)`. Then angular velocity will beA. 1 rad/sB. 2 rad/sC. 4 rad/sD. 5 rad/s

The total energy of a particle executing S.H.M. is proportional toA. square of amplitude of motionB. frequency of oscillationC. velocity in equilibrium positionD. displacement from equilibrium position

The average displacement over a period of S.H.M. is (A = amplitude of S.H.M)

The potential energy of a particle with displacement X is `U(X)`. The motion is simple harmonic, when (K is a positive constant)A. `u=(-kx^(2))/(2)`B. `u=(1)/(2)kx^(2)`C. `u=k`D. `u=kx`

The phase difference between displacement and acceleration of a particle performing S.H.M. isA. `(pi)/(2)`B. `pi` radC. `2pi` radD. `(3pi)/(2)` rad

Why is the term angular frequency (ω) used here for a linear motion ?

The total energy of a particle executing S.H.M. is proportional toA. square of amplitudeB. square root of angular velocityC. amplitudeD. angular velocity

State at which point during an oscillation the oscillator has zero velocity but positive acceleration ?

The period of a simple pendulum is doubled, whenA. its length is doubledB. the mass of the bob is doubledC. its length is four timesD. the amplitude of the pendulum is double and the length of the pendulum also double

What is the effect on the time period of a simple pendulum if the mass of **** is doubled?A) HalvedB) DoubledC) Becomes eight timeD) No effect

What is the nature of thermal changes in air, when a sound wave propagates through it?

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance `(2A)/3` from equilibrium position. The new amplitude of the motion is:A. `(A)/(3)sqrt(41)`B. `3A`C. `Asqrt(3)`D. `(7A)/(3)`

A particle moves in a circular path with a continuously increasing speed. Its motion isA. nonperiodicB. OscillatoryC. Simple harmonicD. UCM

What are the Equation of beats ?

You have a light spring, a metre scale and a known mass. How will you find time period of vibration of mass without the use of clock?

A body of mass 25 gm performs linear S.H.M. The force constant of the motion is 400 dyne/cm. When the body is at a distance of 10 cm from the equilibrium position has velocity of 40 cm/s, then the total energy of the body will beA. `40xx10^(4)J`B. `4xx10^(4)erg`C. `2xx10^(4)erg`D. `2xx10^(5)J`

When a force F1 acts on a particle, frequency is 6 Hz, when a force F2 acts, frequency is 8 Hz. What is the frequency when both the forces act simultaneously in the same direction?

A particle is moving in a circle with uniform speed its motion isA. periodic and simple harmonicB. a periodicC. periodic but not simple harmonicD. non periodic but simple harmonic

In simple harmonic motion, the quantities are not constantA. amplitude and frequencyB. potential energy and kinetic energyC. total energy and propagation constantD. path length

The necessary and sufficient condition for S.H.M. isA. constant period and inertial propertyB. constant acceleration and elasticity propertyC. proportionality between equilibrium position in opposite directionD. periodic and harmonic

The maximum velocity of a point undergoing simultaneous two oscillations given by `x_(1) =a cos wt` and `x_(2) = a cos 2 wt`is:A. `2.74 aw`B. `2aw`C. `1.414aw`D. `aw`

A particle executing SHM has a maximum speed of `30 cm//s` and a maximum acceleration of `60 cm/s^(2)` . The period of oscillation isA. `pisec`B. `(pi)/(2)sec`C. `2pisec`D. `(pi)/(t)sec `