Physics - Oscillations - Quiz-5

A particle in SHM is described by the displacement function $x (t) = A \cos (ω t + ϕ), ω = 2 π / T .$ If the initial (t = 0) position of the particle is 1 cm, its initial velocity is $π c m s^{- 1}$ and its angular frequency is $π s^{- 1}$ , then the amplitude of its motion is:

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$π c m$
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2 cm
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$\sqrt{2} c m$
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1 cm

A particle starts SHM from the mean position. Its amplitude is 'a' and total energy E. At one instant its kinetic energy is 3E/4. Its displacement at this instant is :

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$y = a / \sqrt{2}$
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$y = \frac{a}{2}$
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$y = \frac{a}{\sqrt{3 / 2}}$
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y = a

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic?

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1 cm
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$\sqrt{2} c m$
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3 cm
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$2 \sqrt{2} c m$

A body of mass 500 g is attached to a horizontal spring of spring constant 8 $π^{2} {Nm}^{- 1} .$ If the body is pulled to a distance of 10 cm from its mean position, then its frequency of oscillation is :

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2 Hz
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4 Hz
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8 Hz
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0.5 Hz

Two springs of force constants k and 2k are connected to a mass m as shown below. The frequency of oscillation of the mass is:

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$\frac{1}{2 π} \sqrt{k / m}$
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$\frac{1}{2 π} \sqrt{2 k / m}$
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$\frac{1}{2 π} \sqrt{\frac{3 k}{m}}$
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$\frac{1}{2 π} \sqrt{\frac{m}{k}}$

A weightless spring that has a force constant k oscillates with frequency n when a mass m is suspended from it. The spring is cut into equal halves and a mass 2m is suspended from one part of the spring. The frequency of oscillation will now become:

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n
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2n
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$\frac{n}{\sqrt{2}}$
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$n {(2)}^{1 / 2}$

An object suspended from a spring exhibits oscillations of period T. Now, the spring is cut in two halves and the same object is suspended with two halves as shown in the figure. The new time period of oscillation will become :

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$\frac{T}{2 \sqrt{2}}$
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$\frac{T}{2}$
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$\frac{T}{\sqrt{2}}$
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2T

A mass 1 kg suspended from a spring whose force constant is $400 {Nm}^{- 1}$ , executes simple harmonic oscillation. When the total energy of the oscillator is 2 J, the maximum acceleration experienced by the mass will be:

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$2 {ms}^{- 2}$
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$4 {ms}^{- 2}$
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$40 {ms}^{- 2}$
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$400 {ms}^{- 2}$

What will be the force constant of the spring system shown in the figure?

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$\frac{k_{1}}{2} + k_{2}$
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${[\frac{1}{2 k_{1}} + \frac{1}{k_{2}}]}^{- 1}$
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$\frac{1}{2 k_{1}} + \frac{1}{k_{2}}$
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${[\frac{2}{k_{1}} + \frac{1}{k_{2}}]}^{- 1}$

One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance $x_{0}$ from the block. The spring is then compressed by $2 x_{0}$ and then released. The time taken to strike the wall will be

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$\frac{1}{6} π \sqrt{\frac{k}{m}}$
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$\sqrt{\frac{k}{m}}$
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$\frac{2 π}{3} \sqrt{\frac{m}{k}}$
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$\frac{π}{4} \sqrt{\frac{k}{m}}$

A piece of wood had dimensions a, b and c. Its relative density is d. It is floating in water such that the side c is vertical. It is now pushed down gently and released. The time period is:

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$T = 2 π \sqrt{(\frac{abc}{g})}$
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$T = 2 π \sqrt{(\frac{ba}{dg})}$
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$T = 2 π \sqrt{(\frac{g}{dc})}$
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$T = 2 π \sqrt{(\frac{dc}{g})}$

Which of the following figure represent(s) damped simple harmonic motions?

The amplitude of the damped oscillator becomes (1/3) in 2s. Its amplitude after 6s is 1/n times the original. Then n is equal to:

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$2^{3}$
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$3^{2}$
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$3^{1 / 3}$
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$3^{3}$

The distance covered by a particle undergoing SHM in one time period is: (amplitude=A)

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zero
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A
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2 A
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4 A

The maximum kinetic energy of a simple pendulum is E. The maximum potential energy of pendulum:

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is equal to E.
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is less than or E.
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maybe equal to E.
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is greater than E.

A smooth narrow tunnel is dug along the chord of the earth. The time period of vibration of a small ball dropped from one end of the tunnel is

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60 min
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84.6 min
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42.3 min
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30 min

The time period of vibration of a uniform disc of mass 'M' and radius 'R' about an axis perpendicular to the plane of disc and passing from a point at a distance $\frac{R}{2}$ from the center of the disc is:

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$2 π \sqrt{\frac{3 R}{2 g}}$
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$2 π \frac{3}{\sqrt{2}} \sqrt{\frac{R}{g}}$
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$2 π \sqrt{\frac{2 R}{3 g}}$
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$2 π \frac{3}{2} \sqrt{\frac{R}{g}}$

Two identical blocks each of mass 'M' are connected with springs of spring constant K and placed on a smooth surface as shown in the figure. When the blocks are in contact the springs are in its natural length. The collision between the masses is elastic. The frequency of vibration on disturbing the masses symmetrically in the directions of arrows and releasing them is

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$\frac{1}{2 π} \sqrt{\frac{K}{M}}$
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$\sqrt{\frac{K}{M}}$
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$\frac{1}{4 π} \sqrt{\frac{K}{M}}$
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$\frac{1}{π} \sqrt{\frac{K}{M}}$

Two mutually perpendicular simple harmonic vibrations of the same frequency superimpose on each other. The amplitude of the two vibrations is different and they are different from each other in phase. The resultant of superposition is

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Parabola
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Straight line
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Elliptical
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Circular

A particle under SHM has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s? The time period of oscillation (in second) is

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$\frac{π}{2}$
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2 $π$
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$π$
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4 $π$

Two equations of S.H.M. are $y_{1} = asin (ωt - α)$ and $y_{2} = bcos (ωt - α)$ . The phase difference between the two is:

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$0^{0}$
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$α^{0}$
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$90^{0}$
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$180^{0}$

A ring of radius R is hung by a nail on its periphery such that it can freely rotate in its vertical plane. The time period of the ring for small oscillations is:

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$T = 2 π \sqrt{\frac{R}{g}}$
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$T = π \sqrt{\frac{R}{g}}$
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$T = 2 π \sqrt{\frac{2 R}{g}}$
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$T = 2 π \sqrt{\frac{3 R}{5 g}}$

If the potential energy U (in J) of a body executing S.H.M. is given by U = 20 + 10( $\sin^{2}$ 100 $π$ t), then the minimum potential energy of the body will be

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Zero
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30 J
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20 J
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40 J

The equation of S.H.M. is given as x = Asin(0.02 $πt$ ), where t is in seconds. With what time period the potential energy oscillates?

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200 s
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100 s
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50 s
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10 s

In a stationary lift a spring-block system oscillates with a frequency f. When the lift accelerates up the frequency becomes f', then

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f' > f
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f' < f
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f' = f
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Any of the above depending on the value of the acceleration of the lift

The kinetic energy (K) of a simple harmonic oscillator varies with displacement (x) as shown. The period of the oscillation will be (mass of oscillator is 1 kg)

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$\frac{π}{2}$ sec
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$\frac{1}{2}$ sec
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$π$ sec
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1 sec

The equation of an SHM is given as y=3sinωt + 4cosωt where y is in centimeters. The amplitude of the SHM will be

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3 cm
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3.5 cm
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4 cm
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5 cm

The equation of a SHM is given as $x = 5 \sin (4 πt + \frac{π}{3})$ , where t is in seconds and x

in meters. During a complete cycle, the average speed of the oscillator is:

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Zero
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10 m/s
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20 m/s
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40 m/s

The equation of a simple harmonic oscillator is given as $y = Asin (20 πt + \frac{π}{3})$ , where t is in seconds. The frequency with which kinetic energy oscillates is

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5 Hz
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10 Hz
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20 Hz
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40 Hz

What is the period of oscillation of the block shown in the figure?

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$2 π \sqrt{\frac{M}{k}}$
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$2 π \sqrt{\frac{4 M}{k}}$
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$π \sqrt{\frac{M}{k}}$
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$2 π \sqrt{\frac{M}{2 k}}$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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