Simple Harmonic Motion — Page 9 | JEE Physics

Q81

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both (A) and (R) are true but (R) is not the correct explanation of (A).

B (A) is true but (R) is false.

C Both (A) and (R) are true and (R) is the correct explanation of (A).

D (A) is false but (R) is true.

Correct Answer

Option C

Solution

\begin{aligned} &\text{ As } h \text{ increases, } g \text{ decreases, } \mathrm{T} \text{ increases }\\ &\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\ & \mathrm{~g}=\frac{\mathrm{g}_0 \mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \end{aligned} \end{aligned}

Q82

The period of oscillation of a simple pendulum is

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

. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :

A 1.30%

B 1.33%

C 1.13%

D 1.03%

Correct Answer

Option C

Solution

Given,

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

.... (i) where, time period, T = 1.95 s Length of string, l = 1 m Acceleration due to gravity = g Error in time period,

\Delta

T = 0.01 s = 10 $-$ 2 s Error in length,

\Delta

L = 1 mm = 1 $\times$ 10 $-$ 3 m Squaring Eq. (i) on both sides, we get

{T^2} = 4{\pi ^2}{L \over g}

\Rightarrow g = 4{\pi ^2}{L \over {{T^2}}}

\Rightarrow {{\Delta g} \over g} = {{\Delta L} \over L} + {{2\Delta T} \over T} = {{{{10}^{ - 3}}} \over 1} + {{2 \times {{10}^{ - 2}}} \over {1.95}}

= {10^{ - 3}} + 1.025 \times {10^{ - 2}}

= {10^{ - 3}} + 10.25 \times {10^{ - 3}}

= 11.25 \times {10^{ - 3}}

$\because$

\Delta g/g \times 100 = 11.25 \times {10^{ - 3}} \times {10^2}

= 1.125\% \simeq 1.13\%

Q83

A particle performs simple harmonic motion with amplitude

A.

Its speed is trebled at the instant that it is at a distance

{{2A} \over 3}

from equilibrium position. The new amplitude of the motion is:

A

A\sqrt 3

B

{{7A} \over 3}

C

{A \over 3}\sqrt {41}

D

3A

Correct Answer

Option B

Solution

We know that

V = \omega \sqrt {{A^2} - {x^2}}

Initially

v = \omega \sqrt {{A^2} - {{\left( {{{2A} \over 3}} \right)}^2}}

Finally

3v = \omega \sqrt {A_{new}^2 - {{\left( {{{2A} \over 3}} \right)}^2}}

where

{A_{new}}

= final amplitude (Given at

x = {{2A} \over 3},

velocity to trebled) On dividing we get

{3 \over 1} = {{\sqrt {A_{new}^2 - {{\left( {{{2A} \over 3}} \right)}^2}} } \over {\sqrt {{A^2} - {{\left( {{{2A} \over 3}} \right)}^2}} }}

9\left[ {{A^2} - {{4{A^2}} \over 9}} \right] = A_{new}^2 - {{4{A^2}} \over 9}

$\therefore$

A_{new} = {{7A} \over 3}

Q84

Two simple harmonic motions are represented by the equations

{y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)

and

{y_2} = 0.1\,\cos \,\pi t.

The phase difference of the velocity of particle

1

with respect to the velocity of particle

2

is

A

{\pi \over 3}

B

{{ - \pi } \over 6}

C

{\pi \over 6}

D

{{ - \pi } \over 3}

Correct Answer

Option B

Solution

{v_1} = {{d{y_1}} \over {dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + {\pi \over 3}} \right)

{v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)

$\therefore$ Phase diff.

= {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}

Q85

The period of oscillation of a simple pendulum is

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

. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

A 1 %

B 5 %

C 2 %

D 3 %

Correct Answer

Option D

Solution

Given

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

\Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}

\Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}

[ as

T = {t \over n}

] So, percentage error in

g

=

{{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta t} \over t} \times 100

=

{{0.1} \over {20.0}} \times 100 + 2 \times {1 \over {90}} \times 100

= 2.72 % = 3 %

Q86

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant

t = {T \over 4}s

starting from mean position. Assume that the initial phase of the oscillation is zero.

A 0.62 J

B 6.2

\times

10

-

3 J

C 1.2

\times

103 J

D 6.2

\times

103 J

Correct Answer

Option A

Solution

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

0.2 = 2\pi \sqrt {{{0.5} \over k}}

k = 50 $\pi$ 2 $\approx$ 500 x = A sin ( $\omega$ t + $\phi$ ) = 5 cm sin

\left( {{{\omega T} \over 4} + 0} \right)

= 5 cm sin

\left( {{\pi \over 2}} \right)

= 5 cm

PE = {1 \over 2}k{x^2}

= {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}

= 0.6255

Q87

In a linear Simple Harmonic Motion (SHM) (A) Restoring force is directly proportional to the displacement. (B) The acceleration and displacement are opposite in direction. (C) The velocity is maximum at mean position. (D) The acceleration is minimum at extreme points. Choose the correct answer from the options given below:

A

{\text{(A), (B) and (D) only }}

B (C) and (D) only

C (A), (B) and (C) Only

D (A), (C) and (D) only

Correct Answer

Option C

Solution

The correct options are: (A) Restoring force is directly proportional to the displacement. - True (this is a defining characteristic of SHM) (B) The acceleration and displacement are opposite in direction. - True (the acceleration is proportional to the displacement but in the opposite direction) (C) The velocity is maximum at mean position. - True (the velocity is zero at the extreme positions and reaches a maximum at the mean position) (D) The acceleration is minimum at extreme points. - False (the acceleration is maximum at the extreme points and zero at the mean position)

Q88

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

A 0.77

B 0.57

C 0.37

D 0.17

Correct Answer

Option C

Solution

Initially : After putting 2 masses of each 'm' at a distance

{L \over 2}

from center : We know, Time period (T) = 2 $\pi$

\sqrt {{{\rm I} \over C}}

$\therefore$ T $\propto$

\sqrt {\rm I}

$\therefore$ Frequency (f) $\propto$

\sqrt {{1 \over {\rm I}}}

$\therefore$

{{{f_1}} \over {{f_2}}}

=

\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}

Also given that, After putting two masses 'm' at both end new frequency becomes 80% of initial frequency. $\therefore$ f2 = 0.8f1 $\therefore$

{{{f_1}} \over {0.8{f_1}}}

=

\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}

$\therefore$

{{{{{\rm I}_1}} \over {{{\rm I}_2}}}}

= 0.64 Initial moment of inertia of the system,

{{{\rm I}_1}}

=

{{M{{\left( {2L} \right)}^2}} \over {12}}

Final moment of inertia of the system, I2 =

{{M{{\left( {2L} \right)}^2}} \over {12}}

+ 2

\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)

$\therefore$

{{M{{\left( {2L} \right)}^2}} \over {12}}

= 0.64

\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]

$\Rightarrow$

{{M{L^2}} \over {3 \times 0.64}}

=

{{M{L^2}} \over 3}

+

{{M{L^2}} \over 2}

$\Rightarrow$

{M \over {1.92}} - {M \over 3} = {m \over 2}

$\Rightarrow$

{{1.08M} \over {3 \times 1.92}}

=

{m \over 2}

$\Rightarrow$

{m \over M}

=

{{1.08 \times 2} \over {3 \times 1.92}}

= 0.37

Q89

A mass

M

is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes

SHM

of time period

T.

If the mass is increased by

m.

the time period becomes

{{5T} \over 3}

. Then the ratio of

{{m} \over M}

is

A

{3 \over 5}

B

{25 \over 9}

C

{16 \over 9}

D

{5 \over 3}

Correct Answer

Option C

Solution

The time period of a simple harmonic motion (SHM) performed by a mass-spring system is given by the formula:

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

where: T is the time period, M is the mass of the object, and k is the spring constant.

We know that if the mass is increased by m, the time period becomes $\dfrac{5T}{3}$ .

We can set up an equation for this new scenario:

\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}

Since we know that $T = 2\pi \sqrt{\dfrac{M}{k}}$ , we can substitute T in the equation above:

\frac{5}{3} \cdot 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M + m}{k}}

Squaring both sides of the equation to eliminate the square root, we get:

\frac{25}{9} \cdot \frac{M}{k} = \frac{M + m}{k}

Solving for $\dfrac{m}{M}$ , we get:

\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}

Q90

In forced oscillation of a particle the amplitude is maximum for a frequency

{\omega _1}

of the force while the energy is maximum for a frequency

{\omega _2}

of the force; then

A

{\omega _1} < {\omega _2}

when damping is small and

{\omega _1} > {\omega _2}

when damping is large

B

{\omega _1} > {\omega _2}

C

{\omega _1} = {\omega _2}

D

{\omega _1} < {\omega _2}

Correct Answer

Option C

Solution

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition) $\therefore$

{\omega _1} = {\omega _2}