Simple Harmonic Motion — Page 4 | JEE Physics

Q31

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by y(t) = y0 sin2

\omega

t, where 'y' is measured from the lower end of unstretched spring. Then

\omega

is:

A

\sqrt {{g \over {{y_0}}}}

B

{1 \over 2}\sqrt {{g \over {{y_0}}}}

C

\sqrt {{{2g} \over {{y_0}}}}

D

\sqrt {{g \over {2{y_0}}}}

Correct Answer

Option D

Solution

y(t) = y0 sin2 $\omega$ t =

{1 \over 2}{y_0}\left( {2{{\sin }^2}\omega t} \right)

=

{1 \over 2}{y_0}\left( {1 - \cos 2\omega t} \right)

From comparing standard equation of SHM Amplitude A =

{{{y_0}} \over 2}

And frequency = 2 $\omega$ At equilibrium situation,

{{mg} \over k} = {{{y_0}} \over 2}

$\Rightarrow$

{k \over m} = {{2g} \over {{y_0}}}

$\therefore$ 2 $\omega$ =

\sqrt {{k \over m}}

$\Rightarrow$ 2 $\omega$ =

\sqrt {{{2g} \over {{y_0}}}}

$\Rightarrow$ $\omega$ =

\sqrt {{{2g} \over {4{y_0}}}}

=

\sqrt {{g \over {2{y_0}}}}

Q32

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is :

A 1

B

{1 \over 2}

C

\sqrt 2

D

{1 \over {\sqrt 2 }}

Correct Answer

Option D

Solution

At equilibrium position V0 = V

{V_0} = {\omega _1}A = \sqrt {{K \over m}} A

.....(i)

V = \omega {A^1} = \sqrt {{K \over {{m \over 2}}}} {A^1}

.....(ii) $\therefore$

{A^1} = {A \over {\sqrt 2 }}

Q33

A spring mass system (mass m, spring constant k and natural length

l

) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity

\omega

, (k

\gg m{\omega ^2}

) the relative change in the length of the spring is best given by the option :

A

{{m{\omega ^2}} \over {3k}}

B

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

C

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

D

\sqrt {{2 \over 3}} \left( {{{m{\omega ^2}} \over k}} \right)

Correct Answer

Option B

Solution

m

{\omega ^2}

(l0 + x) = kx x =

{{m{I_0}{\omega ^2}} \over {k - m{\omega ^2}}}

For k >> m

{\omega ^2}

{x \over {{I_0}}} = {{m{\omega ^2}} \over k}

Q34

When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :

A circular

B straight line

C parabolic

D elliptical

Correct Answer

Option D

Solution

Since, the particle is executing SHM. Therefore, displacement equation of wave will be

y = A\sin \omega t

\Rightarrow y/A = \sin \omega t

and wave velocity equation will be

{v_y} = {{dy} \over {dt}} = A\omega \cos \omega t

\Rightarrow {v_y}/A\omega = \cos \omega t

Now,

{\sin ^2}\omega t + {\cos ^2}\omega t = 1

$\therefore$

{(y/A)^2} + {({v_y} / A\omega )^2} = 1

This equation is similar to the equation of ellipse.

Q35

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

A 16 m/s2

B 2

\pi

2 ms

-

2

C

\pi

2 ms

-

2

D 9.8 ms

-

2

Correct Answer

Option B

Solution

The formula for the period of a simple pendulum is given by:

T = 2\pi\sqrt{\frac{l}{g}}

where: T is the period of the pendulum, l is the length of the pendulum, and g is the acceleration due to gravity.

We need to find g.

Rearranging the formula for g, we get:

g = \frac{4\pi^2l}{T^2}

Given: l = 2 m, T = 2 s, Substituting these values into the equation, we get:

g = \frac{4\pi^2 \times 2}{(2)^2} = 2\pi^2 \, ms^{-2}

Q36

Y = A sin(

\omega

t +

\phi

0) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is

Y = {A \over 2}

and it is moving along negative x-direction. Then the initial phase angle

\phi

0 will be:

A

{{5\pi } \over 6}

B

{{\pi } \over 3}

C

{{2\pi } \over 3}

D

{{\pi } \over 6}

Correct Answer

Option A

Solution

The initial phase angle

{\phi _0} = \pi - {\pi \over 6} = {{5\pi } \over 6}

Q37

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period :

A

2\pi \sqrt {{R \over g}}

B

{g \over {2\pi R}}

C

{{2\pi R} \over g}

D

{1 \over {2\pi }}\sqrt {{g \over R}}

Correct Answer

Option A

Solution

Value of g on the particle of mass m,

g = {{GMd} \over {{R^3}}}

Force acting on the particle towards the center of the earth, F = mg Force along the tunnel = F cos $\theta$ = F1 = mg cos $\theta$ = m .

{{GMd} \over {{R^3}}}

\left( {{x \over d}} \right)

=

{{GMm} \over {{R^3}}}

. x =

{{{g_s}m} \over R}.\,x

[as gs =

{{GM} \over {{R^2}}}

at earth surface] $\therefore$ acceleration along the tunnel $\alpha$ =

{{{F_1}} \over m}

=

{{{g_s}} \over R}.x

Time period =

{{2\pi {} } \over \omega } = 2\pi {\sqrt {{x \over \alpha }} }

[as

{\omega ^2} = {\alpha \over x}

]

= 2\pi {\sqrt {{R \over {{g_s}}}} }

Q38

If two similar springs each of spring constant K1 are joined in series, the new spring constant and time period would be changed by a factor :

A

{1 \over 2},2\sqrt 2

B

{1 \over 4},2\sqrt 2

C

{1 \over 2},\sqrt 2

D

{1 \over 4},\sqrt 2

Correct Answer

Option C

Solution

{1 \over {{K_{eq}}}} = {1 \over {{K_1}}} + {1 \over {{K_1}}}

\Rightarrow {K_{eq}} = {{{K_1} \times {K_1}} \over {{K_1} + {K_2}}} = {{K_1^2} \over {2{K_1}}} = {{{K_2}} \over 2}

$\therefore$

T' = 2\pi {\sqrt {{m \over {{K_{eq}}}}} }

= 2\pi {\sqrt {{m \over {{{{K_1}} \over 2}}}} }

= 2\pi {\sqrt {{{2m} \over {{K_1}}}} }

= \sqrt 2 T

[ where

T = 2\pi {\sqrt {{m \over {{K_1}}}} }

]

Q39

A particle executes S.H.M., the graph of velocity as a function of displacement is :

A a parabola

B a helix

C an ellipse

D a circle

Correct Answer

Option C

Solution

For a body performing SHM, relation between velocity and displacement

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

now, square both side

{v^2} = {w^2}({A^2} - {x^2})

\Rightarrow {v^2} = {w^2}{A^2} - {\omega ^2}{x^2}

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

divide whole equation by

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

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

{{{v^2}} \over {{{(\omega A)}^2}}} + {{{x^2}} \over {{{(A)}^2}}} = 1

above equation is similar as standard equation of ellipses, so graph between velocity and displacement will be ellipses.

Q40

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to drop to half of its initial value? (ln 2 = 0.693)

A 17.32 s

B 34.65 s

C 0.034 s

D 15.01 s

Correct Answer

Option B

Solution

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

{{bt} \over {2m}} = \ln 2 = 0.693

t = {{2m} \over b} \times 0.693

t = 2 \times {{500} \over {20}} \times 0.693

t = 50 \times 0.693 = 34.6

sec.