DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-

Statement 1 :The amplitude of an oscillating pendulum decreases gradually with time

Statement 2 :The frequency of the pendulum decreases with time.

(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false

Solution

(d)Statement -1 is true, Statement-2 is false

The amplitude of an oscillating pendulum decreases with time due to friction of air. Its frequency doesn’t depend upon the amplitude \(\left ( T=2\pi \sqrt{\frac{\iota }{g}} \right )\)

DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-

Statement 1 : In simple harmonic motion, the motion is to and fro and periodic

Statement 2 :Velocity of the particle \((v)=\omega \sqrt{k^{2}-x^{2}}\)(where x is the displacement).

(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false

Solution

(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1

S.H.M. is to and fro motion of an object and it is periodic.

\(v=\omega \sqrt{k^{2}-x^{2}}\)

If x= 0, v has maximum value. At x = k, v has minimum velocity. Similarly, when x = – k, v has zero value,all these indicate to and from movement.

DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-

Statement -1 : If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times.

Statement -2 : The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.

(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false

Solution

(b)Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1

DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-

Statement 1:The graph between velocity and displacement for a harmonic oscillator is an ellipse.

Statement -2 :Velocity does not change uniformly with displacement in harmonic motion.

(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false

Solution

(c)Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1

A block is resting on a piston which is moving vertically with S.H.M. of period 1.0 s. At what amplitude of motion will the block and piston separate?

0.2 m
0.25 m
0.3 m
0.35 m

Solution

An object undergoing SHM takes 0.5 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. The period, frequency and amplitude of the motion is :

1s, 1Hz, 25 cm
2s, 1Hz, 50 cm
1s, 2Hz, 25 cm
2s, 2Hz, 50 cm

Solution

Given : T/2 = 0.5 s

∴ T = 1s

Frequency,f = ¹⁄_T = ¹⁄₁ = 1hz

If A is the amplitude, then

2A = 50 cm ⇒ A =25 cm

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

first decrease and then increase to the original value
first increase and then decrease to the original value
increase towards a saturation value
remain unchanged

Solution

Centre of mass of combination of liquid and hollow portion(at position l), first goes down (to l + Δl) and when total water is drained out, centre of mass regain its original position(to l),

T = 2 π \(\sqrt{\frac{l}{g}}\)

∴ ‘T’ first increases and then decreases to original value.

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes.They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

zero
2π/3
π
π/6

Solution

Out of the following functions, representing motion of a particle, which represents S.H.M. ?

y = sin ω t - cos ω t
y = sin³ ω t
y = 5 cos(^3π⁄₄ - 3 ω t)
y = 1 + ω t + ω² t²

Solution

Only functions given in (A) and (C) represent S.H.M.

A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time ?

Solution

The given velocity-position graph depicts that the motion of the particle is S.H.M.

In SHM, at t = 0, v = 0 and x= x_max

So, option (a)is correct.

The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be

T
T/√2
2T
√2T

Solution

The displacement of a particle along the x-axis is given by x = a sin² ω t.The motion of the particle corresponds to

simple harmonic motion of frequency ω/π
simple harmonic motion of frequency 3ω/2π
non simple harmonic motion
simple harmonic motion of frequency ω/2π

Solution

A simple pendulum is suspended from the roof a trolley which moves in a horizontal direction with an accelerationa.Then the time period is given by \(T=2\pi \sqrt{\frac{\iota }{g}}\), where g’ is equal to

g
g - a
g + a
\(\sqrt{g^{2}+a^{2}}\)

Solution

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