Simple Harmonic Motion — Page 2 | NEET Physics

Q11

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is :

A 4n

B n

C 2n

D 3n

Correct Answer

Option C

Solution

Displacement equation of SHM of frequency 'n' x = A sin ( $\omega$ t) = A sin (2 $\pi$ nt) Now, Potential energy

U = {1 \over 2}k{x^2} = {1 \over 2}K{A^2}{\sin ^2}(2\pi nt)

= {1 \over 2}k{A^2}\left[ {{{1 - \cos (2\pi (2n)t)} \over 2}} \right]

So frequency of potential energy = 2n

Q12

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is -

A 0.628 s

B 0.0628 s

C 6.28 s

D 3.14 s

Correct Answer

Option A

Solution

F = Kx 10 = K $\times$ 0.05 K =

{{1000} \over 5} = 200

T = 2\pi \sqrt {{m \over K}} = 2\pi \sqrt {{2 \over {200}}}

= {{2\pi } \over {10}} = {{6.28} \over {10}}

= .628 s

Q13

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :

A

{{3\pi } \over 2}rad

B

{\pi \over 2}rad

C zero

D

\pi

rad

Correct Answer

Option D

Solution

Displacement, y = A.sin $\omega$ t Velocity,

\upsilon

=

{{dy} \over {dt}}A\omega .\cos \omega t

Acceleration, a =

- A{\omega ^2}.\sin \omega t

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

So the phase difference between y and a is $\pi$ .

Q14

The radius of circle, the period of revolution, initial position and sense of revolution are indicated is the figure. y- projection of the radius vector of rotating particle P is :

A y(t) = 4sin

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

, where y in m

B y(t) = 3cos

\left( {{{3\pi t} \over 2}} \right)

, where y in m

C y(t) = 3cos

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

, where y in m

D y(t) = -3cos2

\pi

t, where y in m

Correct Answer

Option C

Solution

Angular velocity, ( $\omega$ ) =

{{2\pi } \over T} = {{2\pi } \over 4} = {\pi \over 2}

rad/s Since, at t = 0, displacement (y) is maximum, so equation will be cosine function. y = a cos $\omega$ t $\Rightarrow$ y = 3 cos

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

Q15

Average velocity of a particle executing SHM in one complete vibration is :

A A

\omega

B

{{A{\omega ^2}} \over 2}

C

{{A\omega } \over 2}

D zero

Correct Answer

Option D

Solution

Since, net displacement in complete cycle ∆y = 0 So, Average velocity = Displacement Time travel =

{{{y_f} - {y_i}} \over T}

=

{{\Delta y} \over T} = {0 \over T} = 0

Q16

The displacement of a particle executing simple harmonic motion is given by y = A 0 + A sin

\omega

t + B cos

\omega

t. Then the amplitude of its oscillation is given by :

A

\sqrt {{A^2} + {B^2}}

B A + B

C A +

\sqrt {{A^2} + {B^2}}

D

\sqrt {A_0^2 + {{\left( {A + B} \right)}^2}}

Correct Answer

Option A

Solution

From the given displacement y = A 0 + A sin $\omega$ t + B cos $\omega$ t.

Let assume, y - A 0 = $\gamma$ $\gamma$ = A sin $\omega$ t + B cos $\omega$ t =

\sqrt {{A^2} + {B^2}} \sin \left( {\omega t + \phi } \right)

which is S.H.M $\therefore$ Resultant amplitude of the particle, =

\sqrt {{A^2} + {B^2}}

Q17

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s –2 at a distance of 5 m from the mean position. The time period of oscillation is

A 2

\pi

s

B

\pi

s

C 2 s

D 1 s

Correct Answer

Option B

Solution

Given, acceleration, a = 20 m/s 2 , and displacement, y = 5m Magnitude of acceleration of a particle moving in a SHM is, |a| =

{\omega ^2}

y; where y is amplitude. $\Rightarrow$ 20 =

{\omega ^2}

(5) $\Rightarrow$

\omega = 2

rad/s Time period of pendulum, T =

{{2\pi } \over \omega } = {{2\pi } \over 2}

= $\pi$ s

Q18

A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is K'. Then they are connected in parallel and force constant is k''. Then k' : k'' is

A 1 : 9

B 1 : 11

C 1 : 14

D 1 : 6

Correct Answer

Option B

Solution

Let us assume, the length of spring be l.

When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths

{l \over 6},{{2l} \over 6}

and

{{3l} \over 6}

with force constant, $\therefore$

{k_1} = {{kl} \over {{l_1}}} = {{kl} \over {l/6}} = 6k

{k_2} = {{kl} \over {{l_2}}} = {{kl} \over {2l/6}} = 3k

{k_3} = {{kl} \over {{l_3}}} = {{kl} \over {3l/6}} = 2k

When connected in series,

{1 \over {k'}} = {1 \over {6k}} + {1 \over {3k}} + {1 \over {2k}} = {{1 + 2 + 3} \over {6k}} = {1 \over k}

\therefore k' = k

When connected in parallel, k'' = 6k + 3k + 2k = 11k

{{k'} \over {k''}} = {k \over {11k}} = {1 \over {11}}

Q19

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

A

{{\sqrt 5 } \over {2\pi }}

B

{{4\pi } \over {\sqrt 5 }}

C

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

D

{{\sqrt 5 } \over \pi }

Correct Answer

Option B

Solution

Given, Amplitude A = 3 cm When particle is at x = 2 cm According to question, magnitude of velocity = acceleration

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

\sqrt {{{\left( 3 \right)}^2} - {{\left( 2 \right)}^2}} = 2\left( {{{2\pi } \over T}} \right)

\sqrt 5 = {{4\pi } \over T} \Rightarrow T = {{4\pi } \over {\sqrt 5 }}

Q20

A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is

A

{3 \over 4}

B

{4 \over 3}

C

{{16} \over 9}

D

{9 \over {16}}

Correct Answer

Option D

Solution

The time period of oscillation is given by

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

{T_1} = 3 = 2\pi \sqrt {m/k}

{T_2} = 5 = 2\pi \sqrt {\left( {{{m + 1} \over k}} \right)}

On dividing :

3/5 = \sqrt {m/\left( {m + 1} \right)}

9/25 = \sqrt {m/\left( {m + 1} \right)}

9m + 9 = 25m

16m = 9 \Rightarrow m = 9/16