Simple Harmonic Motion — Page 4 | NEET Physics

Q31

Two simple harmonic motions of angular frequency 100 and 1000 rad s

-

1 have the same displacement amplitude. The ratio of their maximum acceleration is

A

1:{10^3}

B

1:{10^4}

C

1:10

D

1:{10^2}

Correct Answer

Option D

Solution

Maximum acceleration of a particle in the simple harmonic motion is directly proportional to the square of angular frequency i.e.

a \propto {\omega ^2}

$\therefore$

{{{a_1}} \over {{a_2}}} = {{\omega _1^2} \over {\omega _2^2}} = {{{{\left( {100} \right)}^2}} \over {{{\left( {1000} \right)}^2}}} = {1 \over {100}}

$\Rightarrow$ a 1 : a 2 = 1 : 10 2 .

Q32

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is

A

\pi

B 0.707

\pi

C zero

D 0.5

\pi

Correct Answer

Option D

Solution

Let y = Asin $\omega$ t

{{dy} \over {dt}} = A\omega \cos \omega t = A\omega \sin \left( {\omega t + {\pi \over 2}} \right)

Acceleration =

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

The phase difference between acceleration and velocity is

\pi /2

.

Q33

The particle executing simple harmonic motion has a kinetic energy K 0 cos 2

\omega

t. The maximum values of the potential energy and the total energy are respectively

A K 0 /2 and K 0

B K 0 and 2K 0

C K 0 and K 0

D 0 and 2k 0

Correct Answer

Option C

Solution

Kinetic energy + potential energy = total energy When kinetic energy is maximum, potential energy is zero and vice versa.

$\therefore$ Maximum potential energy = total energy.

0 + K 0 = K 0 (K.E.

+ P.E. = total energy).

Q34

A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take g = 10 m/s 2 ).

A 10.0 cm

B any value less than 12.0 cm

C 4.0 cm

D 8.0 cm

Correct Answer

Option A

Solution

Mass gets detached at the upper extreme position when pan returns to its mean position. At that point,

R = mg - m{\omega ^2}a = 0

, i.e.

g = {\omega ^2}a

\Rightarrow a = g/{\omega ^2} = mg/k

\Rightarrow a = {{2 \times 10} \over {200}}

\left[\because {As = {\omega ^2} = {k \over m}} \right]

$\Rightarrow$ a = 1/10 m = 10 cm

Q35

A particle executes simple harmonic oscillation with an amplitude

a

. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

A T/8

B T/12

C T/2

D T/4

Correct Answer

Option B

Solution

Displacement from the mean position

y = a\sin \left( {{{2\pi } \over T}} \right)t

According to problem y = a/2

a/2 = a\sin \left( {{{2\pi } \over T}} \right)t

\Rightarrow {\pi \over 6} = \left( {{{2\pi } \over T}} \right)t \Rightarrow t = T/12

This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.

Q36

A rectangular block of mass m and area of cross-section A floats in a liquid of density

\rho

. If it is given a small vertical displacement from equilibrium it undergoes with a time period T, then

A

T \propto {1 \over {\sqrt m }}

B

T \propto \sqrt \rho

C

T \propto {1 \over {\sqrt A }}

D

T \propto {1 \over \rho }

Correct Answer

Option C

Solution

Let l be the length of block immersed in liquid as shown in the figure.

When the block is floating, $\therefore$ mg = Al $\rho$ m If the block is given vertical displacement y then the effective restoring force is

F = - \left[ {A\left( {l + y} \right)\rho g - mg} \right]

= - \left[ {A\left( {l + y} \right)\rho g - Al\rho g} \right] = - Al\rho gy

Restoring force =

- \left[ {Al\rho g} \right]y

.

As this F is directed towards its equilibrium position of block, so if the block is left free, it will execute simple harmonic motion.

Here inertia factor = mass of block = m Spring factor =

{A\rho g}

$\therefore$ Time period =

T = 2\pi \sqrt {{m \over {A\rho g}}}

i.e.

T \propto {1 \over {\sqrt A }}

Q37

The circular motion of a particle with constant speed is

A periodic but not simple harmonic

B simple harmonic but not periodic

C period and simple harmonic

D neither periodic not simple harmonic.

Correct Answer

Option A

Solution

In circular motion of a particle with constant speed, particle repeats its motion after a regular interval of time but does not oscillate about a fixed point.

So, motion of particle is periodic but not simple harmonic.

Q38

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is

A 4 Hz

B 3 Hz

C 2 Hz

D 1 Hz

Correct Answer

Option D

Solution

In S.H.M,

{v_{\max }} = A\omega = A\left( {2\pi f} \right)

f = {{{v_{\max }}} \over {2\pi A}} = {{31.4} \over {2\left( {3.14} \right) \times 5}} = 1

Hz per sec.

Q39

Two springs of spring constants k 1 and k 2 are joined in series. The effective spring constant of the combination is given by

A

\sqrt {{k_1}{k_2}}

B (k 1 + k 2 )/2

C k 1 + k 2

D k 1 k 2 /(k 1 + k 2 )

Correct Answer

Option D

Solution

{1 \over {{k_{eq}}}} = {1 \over {{k_1}}} + {1 \over {{k_2}}} \Rightarrow {1 \over {{k_{eq}}}} = {{{k_1} + {k_2}} \over {{k_1}{k_2}}}

\Rightarrow {k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}

Q40

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is

A

{2 \over 3}

E

B

{1 \over 8}

E

C

{1 \over 4}

E

D

{1 \over 2}

E

Correct Answer

Option C

Solution

Potential energy of simple harmonic oscillator =

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

for

y = {a \over 2},P.E = {1 \over 2}m{\omega ^2}{{{a^2}} \over 4}

\Rightarrow P.E = {1 \over 4}\left( {{1 \over 2}m{\omega ^2}{a^2}} \right) = {E \over 4}