Simple Harmonic Motion — Page 3 | JEE Physics

Q21

If a simple pendulum has significant amplitude (up to a factor of

1/e

of original ) only in the period between

t = 0s\,\,to\,\,t = \tau \,s,

then

\tau \,

may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with

b

as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds :

A

{{0.693} \over b}

B

b

C

{1 \over b}

D

{2 \over b}

Correct Answer

Option D

Solution

The equation of motion for the pendulum, suffering retardation

I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right)

where

I = m{\ell ^2}

and

\alpha = {d^2}\theta /d{t^2}

$\therefore$

{{{d^2}\theta } \over {d{t^2}}} = - {g \over \ell }\tan \theta + {{bv} \over \ell }

on solving we get

\theta = {\theta _0}\,{e^{{{bt} \over 2}\sin \left( {\omega t + \phi } \right)}}

According to questions

{{{\theta _0}} \over e} = {\theta _0}{e^{{{ - b\tau } \over 2}}}

$\therefore$

\tau = {2 \over b}

Q22

The amplitude of a damped oscillator decreases to

0.9

times its original magnitude in

5s

. In another

10s

it will decrease to

\alpha

times its original magnitude, where

\alpha

equals

A

0.7

B

0.81

C

0.729

D

0.6

Correct Answer

Option C

Solution

as

\,\,A = {A_0}{e^{{{bt} \over {2m}}}}

(where,

{A_0} =

maximum amplitude) According to the questions, after

5

second,

0.9{A_0} = {A_0}{e^{ - {{b\left( 5 \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

After

10

more second,

A = {A_0}{e^{ - {{b\left( {15} \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

From eqns

(i)

and

(ii)

A = 0.729\,{A_0}

$\therefore$

\alpha = 0.729

Q23

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass

M.

The piston and the cylinder have equal cross sectional area

A

. When the piston is in equilibrium, the volume of the gas is

{V_0}

and its pressure is

{P_0}.

The piston is slightly displaced from the equilibrium position and released,. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frquency

A

{1 \over {2\pi }}\,{{A\gamma {P_0}} \over {{V_0}M}}

B

{1 \over {2\pi }}\,{{{V_0}M{P_0}} \over {{A^2}\gamma }}

C

{1 \over {2\pi }}\,\sqrt {{{A\gamma {P_0}} \over {{V_0}M}}}

D

{1 \over {2\pi }}\,\sqrt {{{M{V_0}} \over {A\gamma {P_0}}}}

Correct Answer

Option C

Solution

{{Mg} \over A} = {P_0}

Mg = {P_0}A\,\,\,\,...\left( 1 \right)

{P_0}V_0^\gamma = P{V^\gamma }

P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}

Let piston is displaced by distance

x

Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right)A = {F_{restoring}}

{P_0}A\left( {1 - {{x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^\gamma }}}} \right) = {F_{restoring}}

\left[ {{x_0} - x \approx {x_0}} \right]

F = - {{\gamma {P_0}Ax} \over {{x_0}}}

$\therefore$ Frequency with which piston executes

SHM.

f = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}A} \over {{x_0}M}}} = {1 \over {2\pi }}\sqrt {{{\gamma {P_0}{A^2}} \over {M{V_0}}}}

Q24

A pendulum made of a uniform wire of cross sectional area

A

has time period

T.

When an additional mass

M

is added to its bob, the time period changes to

{T_{M.}}

If the Young's modulus of the material of the wire is

Y

then

{1 \over Y}

is equal to : (

g=

gravitational

acceleration

)

A

\left[ {1 - {{\left( {{{{T_M}} \over T}} \right)}^2}} \right]{A \over {Mg}}

B

\left[ {1 - {{\left( {{T \over {{T_M}}}} \right)}^2}} \right]{A \over {Mg}}

C

\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}

D

\left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{{Mg} \over A}

Correct Answer

Option C

Solution

As we know, time period,

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

When a additional mass

M

is added then

{T_M} = 2\pi \sqrt {{{\ell + \Delta \ell } \over g}}

{{{T_M}} \over T} = \sqrt {{{\ell + \Delta \ell } \over \ell }}

or,

\,\,{\left( {{{{T_M}} \over T}} \right)^2} = {{\ell + \Delta \ell } \over \ell }

or,

\,\,{\left( {{{{T_M}} \over T}} \right)^2} = 1 + {{Mg} \over {Ay}}

\left[ \, \right.

as

\left. {\Delta \ell = {{Mg\ell } \over {Ay}}\,} \right]

$\therefore$

{1 \over y} = \left[ {{{\left( {{{{T_M}} \over T}} \right)}^2} - 1} \right]{A \over {Mg}}

Q25

A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin

{{\pi t} \over {90}}

. The ratio of kinetic to potential energy of this particle at t = 210 s will be:

A

{1 \over 9}

B 3

C 2

D 1

Correct Answer

Option B

Solution

K =

{1 \over 2}

m

{\omega ^2}

A2cos2 $\omega$ t U =

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

A2 sin2 $\omega$ t

{k \over U}

= cot2 $\omega$ t = cot2

{\pi \over {90}}

(210) =

{1 \over 3}

Hence ratio is 3 (most appropriate)

Q26

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1 . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is

{\pi \over 4}

.

A 500 m/s2

B 500

\sqrt 2 m/

s2

C 750 m/s2

D 750

\sqrt 2

m / s2

Correct Answer

Option B

Solution

Mximum velocity, Vmax = a $\omega$ Maximum acceleration, Amax = a $\omega$ 2 Given that,

{{a{\omega ^2}} \over {a\omega }}

= 10 $\Rightarrow$

\,\,\,

$\omega$ = 10 s $-$ 1 Displacement, x = a sin ( $\omega$ t +

{\pi \over 4}

) at t = 0, displacement x = 5

\therefore\,\,\,

5 = a sin

\left( {{\pi \over 4}} \right)

$\Rightarrow$

\,\,\,

5 = a $\times$

{1 \over {\sqrt 2 }}

$\Rightarrow$

\,\,\,

a = 5

{\sqrt 2 }

\therefore\,\,\,

Maximum acceleration, Amax = a $\omega$ 2 = 5

{\sqrt 2 }

$\times$ (10)2 = 500

{\sqrt 2 }

m/s2

Q27

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :

A

{1 \over 4}Hz

B

{1 \over {2\sqrt 2 }}Hz

C

{1 \over 2}Hz

D

2

Hz

Correct Answer

Option C

Solution

For 1 kg block : Here frequency of spring (f) =

{1 \over {2\pi }}\sqrt {{k \over m}}

Given that, F = 1 Hz

\therefore\,\,\,

{1 \over {2\pi }}\sqrt {{k \over 1}}

= 1 $\Rightarrow$

\,\,\,

k = 4 $\pi$ 2 N m $-$ 1 For 8 kg Block : Here two identical springs are attached in parallel. So, Keq = k + k = 2k

\therefore\,\,\,

Frequency of 8 kg block, F' =

{1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}}

=

{1 \over {2\pi }}\sqrt {{{2k} \over 8}}

=

{1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}}

=

{1 \over {2\pi }} \times \pi

=

{1 \over 2}

Hz

Q28

A simple harmonic motion is represented by : y = 5 (sin 3

\pi

t +

\sqrt 3

cos 3

\pi

t) cm The amplitude and time period of the motion are :

A 10 cm,

{3 \over 2}

s

B 5 cm,

{2 \over 3}

s

C 5 cm,

{3 \over 2}

s

D 10 cm,

{2 \over 3}

s

Correct Answer

Option D

Solution

y = 5[sin(3 $\pi$ t) +

\sqrt 3

cos(3 $\pi$ t)] = 10sin

\left( {3\pi t + {\pi \over 3}} \right)

Amplitude = 10 cm T =

{{2\pi } \over w}

=

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

=

{2 \over 3}

sec

Q29

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency

\omega

. If the radius of the bottle is 2.5 cm then

\omega

is close to – (density of water = 103 kg/m3).

A 2.50 rad s

-

1

B 3.75 rad s

-

1

C 5.00 rad s

-

1

D 7.90 rad s

-

1

Correct Answer

Option D

Solution

Restoring force due to pressing the bottle with small amount x, F =

- \left( {\rho Ax} \right)g

$\Rightarrow$ ma =

- \left( {\rho Ax} \right)g

$\Rightarrow$ a =

- \left( {{{\rho Ag} \over m}} \right)x

$\therefore$

{{\omega ^2} = {{\rho Ag} \over m}}

=

{{{\rho \left( {\pi {r^2}} \right)g} \over m}}

$\Rightarrow$ $\omega$ =

\sqrt {{{{{10}^3} \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2} \times 10} \over {310 \times {{10}^{ - 3}}}}}

= 7.90 rad/s

Q30

A simple pendulum is being used to determine th value of gravitational acceleration g at a certain place. Th length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

A 4.40%

B 3.40%

C 2.40%

D 5.40%

Correct Answer

Option A

Solution

T =

2\pi \sqrt {{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}

=

{{0.1} \over {25}} + 2{1 \over {50}}

=

{{11} \over {250}}

$\therefore$ % accuracy =

{{11} \over {250}}

$\times$ 100% = 4.40 %