Simple Harmonic Motion — Page 7 | NEET Physics

Q61

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)

Q62

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.

Q63

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

Q64

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}

Q65

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)

Q66

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

Q67

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}

Q68

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\%

Q69

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}

Q70

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}