A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $$x=\frac{A}{2}$$ and it moves along positive x-axis. The displacement of particle in time t is $$x =

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$. If we substitute these initial conditions into the equation for the displacement of the particle: $x = A \sin(wt + \delta)$ We have: $\frac{A}{2} = A \sin(\delta)$ Dividing both sides by $A$ gives us: $\frac{1}{2} = \sin(\delta)$ The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees). Therefore, the correct answer is $\delta = \frac{\pi}{6}$.

Simple Harmonic Motion Questions — JEE Physics

Q1

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :

A Circular

B Elliptical

C Sinusoidal

D Straight line

Correct Answer

Option B

Solution

Let

x = A\sin \omega t

\Rightarrow v = A\omega \cos \omega t

\Rightarrow v = \, \pm \,\omega \sqrt {{A^2} - {x^2}}

\Rightarrow {{{v^2}} \over {{\omega ^2}}} + {x^2} = {A^2}

$\Rightarrow$ Ellipse

Q2

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is

x=\dfrac{A}{2}

and it moves along positive x-axis. The displacement of particle in time t is

x = A\sin (wt + \delta )

, then the value of

\delta

will be

A

\dfrac{\pi}{2}

B

\dfrac{\pi}{3}

C

\dfrac{\pi}{4}

D

\dfrac{\pi}{6}

Correct Answer

Option D

Solution

The initial condition states that the particle is at position $x=\dfrac{A}{2}$ at $t=0$ .

If we substitute these initial conditions into the equation for the displacement of the particle: $x = A \sin(wt + \delta)$ We have: $\dfrac{A}{2} = A \sin(\delta)$ Dividing both sides by $A$ gives us: $\dfrac{1}{2} = \sin(\delta)$ The angle whose sine is $\dfrac{1}{2}$ is $\delta = \dfrac{\pi}{6}$ radians (or 30 degrees).

Therefore, the correct answer is $\delta = \dfrac{\pi}{6}$ .

Q3

A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :

A

{A \over 2}

B

{A \over {2\sqrt 2 }}

C

{A \over {\sqrt 2 }}

D A

Correct Answer

Option C

Solution

Total energy of particle =

{1 \over 2}k{A^2}

Potential energy (v) =

{1 \over 2}

kx2 Kinetic energy (K) =

{1 \over 2}

kA2 $-$

{1 \over 2}

kx2 According to the question, Potential energy = Kinetic energy $\therefore$

{1 \over 2}

kx2 =

{1 \over 2}

kA2 $-$

{1 \over 2}

kx2 $\Rightarrow$ kx2 =

{1 \over 2}

kA2 $\Rightarrow$ x = $\pm$

{A \over {\sqrt 2 }}

Q4

The displacement of a damped harmonic oscillator is given by x(t ) = e–0.1t cos (10

\pi

t + f). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :

A 27 s

B 13 s

C 7 s

D 4 s

Correct Answer

Option C

Solution

Amplitude at (t = 0) A0 = e–0.1× 0 = 1 $\therefore$

at\,t = t

if

A = {{{A_0}} \over 2}

$\Rightarrow$

{1 \over 2} = {e^{ - 0.1t}}

t = 10 ln 2 = 7 s

Q5

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then :

A

{K_2}

=

{{{K_1}} \over 2}

B K2 = 2K1

C K2 = K1

D K2 =

{{{K_1}} \over 4}

Correct Answer

Option B

Solution

Maximum kinetic energy at lowest point B is given by K = mg

\ell

(1 $-$ cos $\theta$ ) where $\theta$ = angular amp. K1 = mg

\ell

(1 $-$ cos $\theta$ ) K2 = mg(2

\ell

) (1 $-$ cos $\theta$ ) K2 = 2K1.

Q6

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?

A x =

{A \over 2}

B x =

\pm

A

C x =

\pm

{A \over {\sqrt 2 }}

D x = 0

Correct Answer

Option C

Solution

KE = PE

{1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}

$\Rightarrow$

{A^2} - {X^2} = {X^2}

$\Rightarrow$

2{X^2} = {A^2}

$\Rightarrow$

{X^2} = {{{A^2}} \over {\sqrt 2 }}

$\Rightarrow$

X = \pm {A \over {\sqrt 2 }}

Q7

Two particles A and B of equal masses are suspended from two massless springs of spring constants K1 and K2 respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

A

{{{K_1}} \over {{K_2}}}

B

\sqrt {{{{K_1}} \over {{K_2}}}}

C

{{{K_2}} \over {{K_1}}}

D

\sqrt {{{{K_2}} \over {{K_1}}}}

Correct Answer

Option D

Solution

$\because$

{V_{\max }} = A\omega

Given,

{\omega _1}{A_1} = {\omega _2}{A_2}

We know that

\omega = \sqrt {{K \over m}}

$\therefore$

\sqrt {{{{k_1}} \over m}} {A_1} = \sqrt {{{{k_2}} \over m}} {A_2}

$\Rightarrow$

{{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}}

Q8

A particle of mass

m

executes simple harmonic motion with amplitude a and frequency

v.

The average kinetic energy during its motion from the position of equilibrium to the end is

A

2{\pi ^2}\,m{a^2}{v^2}

B

{\pi ^2}\,m{a^2}{v^2}

C

{1 \over 4}\,m{a^2}{v^2}

D

4{\pi ^2}m{a^2}{v^2}

Correct Answer

Option B

Solution

KEY CONCEPT : The instantaneous kinetic energy of a particle executing

S.H.M.

is given by

K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t

$\therefore$ average

K.E. = < K > = < {1 \over 2}m{\omega ^2}{a^2}{\sin ^2}\omega t >

= {1 \over 2}m\omega {}^2{a^2} < {\sin ^2}\omega t >

= {1 \over 2}m{\omega ^2}{a^2}\left( {{1 \over 2}} \right)

\left( \, \right.

as

\left. { < {{\sin }^2}\theta > = {1 \over 2}} \right)

= {1 \over 4}m{\omega ^2}{a^2} = {1 \over 4}m{a^2}{\left( {2\pi v} \right)^2}

\left( \, \right.

\left. {\omega = 2\pi v} \right)

or,

\,\,\,\,\, < K > = {\pi ^2}m{a^2}{v^2}

Q9

If a spring has time period

T,

and is cut into

n

equal parts, then the time period of each part will be

A

T\sqrt n

B

T/\sqrt n

C

nT

D

T

Correct Answer

Option B

Solution

Let the spring constant of the original spring be

k.

Then its time period

T = 2\pi \sqrt {{m \over k}}

where

m

is the mass of oscillating body. When the spring is cut into

n

equal parts, the spring constant of one part becomes

nk.

Therefore the new time period,

T' = 2\pi \sqrt {{m \over {nk}}} = {T \over {\sqrt n }}

Q10

A body executes simple harmonic motion. The potential energy

(P.E),

the kinetic energy

(K.E)

and total energy

(T.E)

are measured as a function of displacement

x.

Which of the following statements is true ?

A

K.E

is maximum when

x=0

B

T.E

is zero when

x=0

C

K.E

is maximum when

x

is maximum

D

P.E

is maximum when

x=0

Correct Answer

Option A

Solution

K.E. = {1 \over 2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)

When

x=0,

K.E

is maximum and is equal to

{1 \over 2}m{\omega ^2}{a^2}.