Simple Harmonic Motion — Page 5 | NEET Physics

Q41

The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be

A T/4

B T

C T/2

D 2T

Correct Answer

Option C

Solution

Let k be the force constant of spring. If k' is the force constant of each part, then

{1 \over l} = {4 \over {k'}} \Rightarrow k' = 4k

$\therefore$ Time period =

2\pi \sqrt {{m \over {4k}}} = {1 \over 2} \times 2\pi \sqrt {{m \over k}} = {T \over 2}

Q42

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?

A When v is maximum,

a

is maximum

B Value of

a

is zero, whatever may be the value of v.

C When v is zero,

a

is zero

D When v is maximum,

a

is zero.

Correct Answer

Option D

Solution

In simple harmonic motion velocity

= A\omega \sin \left( {\omega t + \pi /2} \right)

acceleration

= A{\omega ^2}\sin \left( {\omega t + \pi } \right)

from this we can easily find out that when v is maximum, then

a

is zero.

Q43

In case of a forced vibration, the resonance peak becomes very sharp when the

A damping force is small

B restoring force is small

C applied periodic force is small

D quality factor is small

Correct Answer

Option A

Solution

The resonance wave becomes very sharp when damping force is small.

Q44

Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is

A

\pm

a

/2

B +

a

C

\pm

a

D

-

1

Correct Answer

Option C

Solution

For a simple harmonic motion between A and B, with O as the mean position, maximum kinetic energy of the particle executing SHM will be at O and maximum potential energy will be at A and B.

$\therefore$ Displacement between maximum potential energy and maximum kinetic energy is

\pm a

.

Q45

When an oscillator completes 100 oscillations its amplitude reduced to

{1 \over 3}

of initial value. What will be its amplitude, when it complettes 200 oscillations ?

A

{1 \over 8}

B

{2 \over 3}

C

{1 \over 6}

D

{1 \over 9}

Correct Answer

Option D

Solution

This is a case of damped vibration as the amplitude of vibration is decreasing with time.

Amplitude of vibrations at any instant t is given by

a = a_0e^{–bt}

, where

a_0

is the initial amplitude of vibrations and b is the damping constant. Now, when t = 100T,

a = a_0/3

[T is time period] Let the amplitude be

a'

at t = 200T. i.e. after completing 200 oscillations. $\therefore$

a = a_0/3 = a0e^{–100Tb}

...(i) and

a' = a_0e^{–200Tb}

...(ii) From (i),

{1 \over 3} = {e^{ - 100Tb}}

$\therefore$

{e^{ - 200Tb}} = 1/9

From (ii),

a' = {a_0} \times {1 \over 9} = {{{a_0}} \over 9}

$\therefore$ The amplitude will be reduced to 1/9 of initial value.

Q46

A mass is suspended separately by two different springs in (successive order then time periods is t 1 and t 2 respectively, If it is connected by both spring as shown in figure then time period is t 0 , the correct relation is

A

t_0^2 = t_1^2 + t_2^2

B

t_0^{ - 2} = t_1^{ - 2} + t_2^{ - 2}

C

t_0^{ - 1} = t_1^{ - 1} + t_2^{ - 1}

D

{t_0} = {t_1} + {t_2}

Correct Answer

Option B

Solution

The time period of a spring mass system as shown in figure 1 is given by

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

, where k is the spring constant. $\therefore$

{t_1} = 2\pi \sqrt {m/{k_1}}

...(i) and

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

...(ii) Now, when they are connected in parallel as shown in figure 2(a), the system can be replaced by a single spring of spring constant, k eff = k 1 + k 2 .

[Since mg = k 1 x + k 2 x = k eff x] $\therefore$

{t_0} = 2\pi \sqrt {m/{k_{eff}}} = 2\pi \sqrt {m/\left( {{k_1} + {k_2}} \right)}

...(iii) From (i),

{1 \over {t_1^2}} = {1 \over {4{\pi ^2}}} \times {{{k_1}} \over m}

...(iv) From (ii),

{1 \over {t_2^2}} = {1 \over {4{\pi ^2}}} \times {{{k_2}} \over m}

...(v) From (ii),

{1 \over {t_0^2}} = {1 \over {4{\pi ^2}}} \times {{{k_1} + {k_2}} \over m}

...(vi) Now (iv) + (v)

{1 \over {t_1^2}} + {1 \over {t_2^2}} = {1 \over {4{\pi ^2}m}}\left( {{k_1} + {k_2}} \right) = {1 \over {t_0^2}}

$\therefore$

t_0^{ - 2} = t_1^{ - 2} + t_2^{ - 2}

Q47

The total energy of particle performing SHM depend on

A k, a, m

B k, a

C k, a, x

D k, x

Correct Answer

Option B

Solution

Energy =

{1 \over 2}m{\omega ^2}{a^2} = {1 \over 2}k{a^2}

Q48

Two masses

M

A and

M

B are hung from two strings of length

l

A and

l

B respectively. They are executing SHM with frequency relation

f

A = 2

f

B , then relation

A

{l_A} = {{{l_B}} \over 4}

does not depend on mass

B

{l_A} = 4{l_B}

, does not depend on mass

C

{l_A} = 2{l_B}

and

{M_A} = 2{M_B}

D

{l_A} = {{{l_B}} \over 2}

and

{M_A} = {{{M_B}} \over 2}

.

Correct Answer

Option A

Solution

{f_A} = 2{f_B}

\Rightarrow {1 \over {2\pi }}\sqrt {{g \over {{l_A}}}} = 2 \times {1 \over {2\pi }}\sqrt {{g \over {{l_B}}}}

\Rightarrow {l \over {{l_A}}} = 4 \times {l \over {{l_B}}}

\Rightarrow {l_A} = {{{l_B}} \over 4}

, which does not depend on mass.

Q49

The bob of simple pendulum having length

l

, is displaced from mean position to an angular position q with respect to vertical. If it is released, then velocity of bob at equilibrium position

A

\sqrt {2gl\left( {1 - \cos \theta } \right)}

B

\sqrt {2gl\left( {1 + \cos \theta } \right)}

C

\sqrt {2gl\cos \theta }

D

\sqrt {2gl}

Correct Answer

Option A

Solution

In

\Delta OAC,\,\cos \theta = A/l

\Rightarrow OA = lcos\theta

$\therefore$

AB = l\left( {1 - \cos \theta } \right) = h

At point, C the velocity of bob = 0. The vertical acceleration = g $\therefore$

{v^2} = 2gh

\Rightarrow v = \sqrt {2gl\left( {1 - \cos \theta } \right)}

Q50

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :

A Circular

B Elliptical

C Sinusoidal

D Straight line

Correct Answer

Option B

Solution

Let

x = A\sin \omega t

\Rightarrow v = A\omega \cos \omega t

\Rightarrow v = \, \pm \,\omega \sqrt {{A^2} - {x^2}}

\Rightarrow {{{v^2}} \over {{\omega ^2}}} + {x^2} = {A^2}

$\Rightarrow$ Ellipse