Simple Harmonic Motion — Page 8 | JEE Physics

Q71

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

Q72

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

Q73

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 $\theta_1 \ell_2 = \theta_2 \ell_1$ .

Q74

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.

Q75

The displacement of particle varies according to the relation

x=4

\left( {\cos \,\pi t + \sin \,\pi t} \right).

The amplitude of the particle is

A

-4

B

4

C

4\sqrt 2

D

8

Correct Answer

Option C

Solution

x = 4\left( {\cos \pi t + \sin \pi t} \right)

= \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)

x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)

Q76

Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is

A

\sqrt{\dfrac{k_2}{k_1}}

B

\sqrt{\dfrac{k_1}{k_2}}

C

\dfrac{k_2}{k_1}

D

\dfrac{k_1}{k_2}

Correct Answer

Option B

Solution

We know, for an oscillation,

{V_{\max }} = wA

.... (1) where, A = amplitude w = frequency of oscillation and

w = \sqrt {{k \over m}}

.... (2) from (1) and (2),

{V_{\max }} = \sqrt {{k \over m}} A

\Rightarrow {V_{\max }} \propto \sqrt k

(As m and A are same (or constant))

\Rightarrow {{{{\left( {{V_{\max }}} \right)}_A}} \over {{{\left( {{V_{\max }}} \right)}_B}}} = \sqrt {{{{k_1}} \over {{k_2}}}}

Hence, option 2 is correct.

Q77

A particle at the end of a spring executes

S.H.M

with a period

{t_1}

. While the corresponding period for another spring is

{t_2}

. If the period of oscillation with the two springs in series is

T

then

A

{T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}

B

{T^2} = t_1^2 + t_2^2

C

T = {t_1} + {t_2}

D

{T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}

Correct Answer

Option B

Solution

For first spring,

{t_1} = 2\pi \sqrt {{m \over {{k_1}}}} ,

For second spring,

{t_2} = 2\pi \sqrt {{m \over {{k_2}}}}

when springs are in series then,

{k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}

$\therefore$

T = 2\pi \sqrt {{{m\left( {{k_1} + {k_2}} \right)} \over {{k_1}{k_2}}}}

$\therefore$

T = 2\pi \sqrt {{m \over {{k_2}}} + {m \over {{k_1}}}}

= 2\pi \sqrt {{{t_2^2} \over {{{\left( {2\pi } \right)}^2}}} + {{t_1^2} \over {{{\left( {2\pi } \right)}^2}}}}

\Rightarrow {T^2} = t_1^2 + t_2^2

Q78

The bob of a pendulum was released from a horizontal position. The length of the pendulum is

10 \mathrm{~m}

. If it dissipates

10 \%

of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use,

\mathrm{g}: 10 \mathrm{~ms}^{-2}

]

A

5 \sqrt{6} \mathrm{~ms}^{-1}

B

5 \sqrt{5} \mathrm{~ms}^{-1}

C

2 \sqrt{5} \mathrm{~ms}^{-1}

D

6 \sqrt{5} \mathrm{~ms}^{-1}

Correct Answer

Option D

Solution

\ell=10 \mathrm{~m}

, Initial energy

=\mathrm{mg} \ell

\begin{aligned} & \text{ So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \\ & \mathrm{v}^2=180 \\ & \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{aligned}

Q79

A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The freqquency of the oscillation is :

A

{1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)

B

{1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)

C

{1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)

D

{1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)

Correct Answer

Option A

Solution

In general equation of simple harmonic motion, y = A sin $\omega$ t

\therefore\,\,\,

a = A sin $\omega$ t0

\,\,\,\,\,\,

b = A sin 2 $\omega$ t0

\,\,\,\,\,\,\,

c = A sin 3 $\omega$ t0 a + c = A[sin $\omega$ t0 + sin 3 $\omega$ t0] = 2A sin 2 $\omega$ t0 cos $\omega$ t0 $\Rightarrow$

\,\,\,

a + c = 2 b cos $\omega$ t0 $\Rightarrow$

\,\,\,

{{a + c} \over b}

= 2 cos $\omega$ t0 $\Rightarrow$

\,\,\,

$\omega$ =

{1 \over {{t_0}}}

cos $-$ 1

\left( {{{a + c} \over {2b}}} \right)

\therefore\,\,\,

f =

{\omega \over {2\pi }}

=

{1 \over {2\pi {t_0}}}

cos $-$ 1

\left( {{{a + c} \over {2b}}} \right)

Q80

A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion . (take ln 2 = 0.693)

A 0.69

\times

102 kg s

-

1

B 3.3

\times

102 kg s

-

1

C 1.16

\times

10

-

2 kg s

-

1

D 5.7

\times

10

-

3 kg s

-

1

Correct Answer

Option C

Solution

A = {A_o}{e^{{{ - b} \over {2m}}t}}

$\Rightarrow$

6 = 12{e^{{{ - b} \over {2 \times 1}} \times 120}}

$\Rightarrow$

6 = 12{e^{ - b \times 60}}

$\Rightarrow$

{1 \over 2} = {e^{ - 60b}}

$\Rightarrow$

\ln (2) = 60b

$\Rightarrow$

b = {{\ln (2)} \over {60}} = 1.16 \times {10^2}

Kg/s