Simple Harmonic Motion — Page 6 | NEET Physics

Q51

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}}}}

Q52

Starting from the origin a body oscillates simple harmonically with a period of

2

s.

After what time will its kinetic energy be

75\%

of the total energy?

A

{1 \over 6}s

B

{1 \over 4}s

C

{1 \over 3}s

D

{1 \over 12}s

Correct Answer

Option A

Solution

K.E.\,

of a body undergoing

SHM

is given by,

K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,

T.E. = {1 \over 2}m{a^2}{\omega ^2}

Given

K.E.=0.75T.E.

\Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}

\Rightarrow t = {\pi \over {6 \times \omega }} \Rightarrow t = {{\pi \times 2} \over {6 \times 2\pi }} \Rightarrow t = {1 \over 6}s

Q53

The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :

A 6 s

B 8 s

C 12 s

D 36 s

Correct Answer

Option D

Solution

Time taken by the harmonic oscillator to move from mean position to half of amplitude is

{T \over {12}}

So,

{T \over {12}}

= 3 T = 36 sec.

Q54

A simple pendulum doing small oscillations at a place

R

height above earth surface has time period of

T_1=4 \mathrm{~s}

.

\mathrm{T}_2

would be it's time period if it is brought to a point which is at a height

2 \mathrm{R}

from earth surface. Choose the correct relation [

\mathrm{R}=

radius of earth] :

A

3 \mathrm{~T}_1=2 \mathrm{~T}_2

B

\mathrm{T}_1=\mathrm{T}_2

C

2 \mathrm{~T}_1=3 \mathrm{~T}_2

D

2 \mathrm{~T}_1=\mathrm{T}_2

Correct Answer

Option A

Solution

The time period of a simple pendulum is given by the formula:

T = 2\pi \sqrt{\frac{l}{g}}

where

T

is the time period,

l

is the length of the pendulum, and

g

is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface.

The acceleration due to gravity at a height

h

above the Earth's surface can be expressed as:

g' = g \left(\frac{R}{R + h}\right)^2

where

g

is the acceleration due to gravity at the surface of the Earth,

R

is the radius of the Earth, and

h

is the height above the Earth’s surface.

Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in

g

due to a change in height will affect the time period. Given that the time period of the pendulum at a height

R

above Earth's surface is

T_1

, and we're to find the time period

T_2

at a height of

2R

, we can use the formula for acceleration due to gravity at different heights to express the relationship between

T_1

and

T_2

. For the initial case at height

R

:

g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}

For the new case at height

2R

:

g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}

The time period is proportional to the square root of the inverse of

g

, so:

\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}

Therefore:

T_1 = \frac{2}{3}T_2

Rearranging this equation:

3T_1 = 2T_2

This corresponds to Option A.

Q55

The maximum velocity of a particle, executing simple harmonic motion with an amplitude

7

mm,

is

4.4

m/s.

The period of oscillation is

A

0.01

s

B

10

s

C

0.1

s

D

100

s

Correct Answer

Option A

Solution

Maximum velocity,

{v_{\max }} = a\omega ,\,\,\,\,\,{v_{\max }} = a \times {{2\pi } \over T}

\Rightarrow T = {{2\pi a} \over {{v_{\max }}}} = {{2 \times 3.14 \times 7 \times {{10}^{ - 3}}} \over {4.4}} \approx 0.01\,s

Q56

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has displacement A while the other one has displacement

{{ - A} \over 2}

and they are moving towards each other. If they cross each other at time t, then t is :

A

{T \over 6}

B

{5T \over 6}

C

{T \over 3}

D

{T \over 4}

Correct Answer

Option A

Solution

Angular displacement ( $\theta$ 1) of particle 1. from equilibrium,

{y_1}

= A sin $\theta$ 1 $\Rightarrow$ A = Asin $\theta$ 1 $\Rightarrow$ sin $\theta$ 1 = 1 = sin

{\pi \over 2}

$\therefore$ $\theta$ 1 =

{\pi \over 2}

Similarly for particle 2 angular displacement $\theta$ 2 from equilibrium, y2 = Asin $\theta$ 2 $\Rightarrow$ $-$

{A \over 2}

= Asin $\theta$ 2 $\Rightarrow$ sin $\theta$ 2 = $-$

{1 \over 2}

= sin

\left( { - {\pi \over 3}} \right)

$\Rightarrow$ $\theta$ 2 = $-$

{{\pi \over 3}}

Relative angular displacement of the two particle, $\theta$ = $\theta$ 1 $-$ $\theta$ 2 =

{{\pi \over 2}}

$-$

\left( { - {\pi \over 6}} \right)

=

{{{2\pi } \over 3}}

Relative angular velocity $=$

\omega - \left( { - \omega } \right)

=

2\omega

If they cross each other at time t then, t =

{\theta \over {2\omega }}

=

{{2\pi } \over {3 \times 2\omega }}

=

{\pi \over {3 \times {{2\pi } \over T}}}

=

{T \over 6}

Q57

A particle executes simple harmonic motion between

x=-A

and

x=+A

. If time taken by particle to go from

x=0

to

\dfrac{A}{2}

is 2 s; then time taken by particle in going from

x=\dfrac{A}{2}

to A is

A 4 s

B 1.5 s

C 3 s

D 2 s

Correct Answer

Option A

Solution

$x=A \sin (\omega t)$

\begin{aligned} & x=\frac{A}{2}=A \sin (\omega t) \\\\ & \frac{1}{2}=\sin (\omega t) \\\\ & t=\left(\frac{\pi}{6 \omega}\right)=2 \\\\ & \frac{\pi}{\omega}=12 \sec \end{aligned}

$x=A=A \sin (\omega t)$ $\omega t=\left(\dfrac{\pi}{2}\right)$ $t=\left(\dfrac{\pi}{2 \omega}\right)=6$ second time $=6-2=4$ seconds

Q58

Given below are two statements : Statement I : A second's pendulum has a time period of 1 second. Statement II : It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are true

D Statement I is true but Statement II is false

Correct Answer

Option B

Solution

As we know time period of second’s penduklum is 2 sec, so statement (1) is incorrect.

Time taken by particle performing SHM between two extreme position is half of the time period.

Here, T = 2 sec.

So, time = 2/2 = 1 sec

Q59

Two simple pendulums having lengths

l_1

and

l_2

with negligible string mass undergo angular displacements

\theta_1

and

\theta_2

, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

A

\theta_1 l_2=\theta_2 l_1

B

\theta_1 l_1=\theta_2 l_2

C

\theta_1 l_2^2=\theta_2 l_1^2

D

\theta_1 l_1^2=\theta_2 l_2^2

Correct Answer

Option A

Solution

Angular Frequency: The angular frequency ( $\omega$ ) of a simple pendulum is given by: $\omega = \sqrt{\dfrac{g}{\ell}}$ where $g$ is the acceleration due to gravity and $\ell$ is the pendulum length.

Angular Acceleration: The angular acceleration ( $\alpha$ ) can be expressed in terms of angular displacement ( $\theta$ ) and angular frequency: $\alpha = -\omega^2 \theta$ Equating Angular Accelerations: Since the angular accelerations of the two pendulums are equal, we equate them: $\dfrac{g}{\ell_1} \theta_1 = \dfrac{g}{\ell_2} \theta_2$ Simplifying the Expression: By canceling out $g$ on both sides, we derive: $\theta_1 \ell_2 = \theta_2 \ell_1$ Thus, the correct expression that relates the displacements and lengths of the pendulums is $\theta_1 \ell_2 = \theta_2 \ell_1$ .

Q60

A child swinging on a swing in sitting position, stands up, then the time period of the swing will

A increase

B decrease

C remains same

D increases of the child is long and decreases if the child is short

Correct Answer

Option B

Solution

KEY CONCEPT : The time period

T = 2\pi \sqrt {{\ell \over g}}

where

\ell

$=$ distance between the point of suspension and the center of mass of the child.

This distance decreases when the child stands $\therefore$

T' < T

i.e., the period decreases.