Simple Harmonic Motion — Page 10 | JEE Physics

Q91

The length of a simple pendulum executing simple harmonic motion is increased by

21\%

. The percentage increase in the time period of the pendulum of increased length is

A

11\%

B

21\%

C

42\%

D

10\%

Correct Answer

Option D

Solution

The period of a simple pendulum is given by:

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

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

Since g is constant, we see that the period T is proportional to the square root of the length L.

If L is increased by 21%, the new length L' is L + 21%L = 1.21L.

The new period T' is then:

T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.21L}{g}} = \sqrt{1.21}T \approx 1.1T

The percentage increase in the time period is then:

\frac{T' - T}{T} \times 100\% = (\sqrt{1.21} - 1) \times 100\% \approx 10\%

Therefore, the percentage increase in the time period of the pendulum of increased length is approximately 10%.

Q92

A particle of mass

m

is attached to a spring (of spring constant

k

) and has a natural angular frequency

{\omega _0}.

An external force

F(t)

proportional to

\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)

is applied to the oscillator. The time displacement of the oscillator will be proportional to

A

{1 \over {m\left( {\omega _0^2 + {\omega ^2}} \right)}}

B

{1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}

C

{m \over {\omega _0^2 - {\omega ^2}}}

D

{m \over {\omega _0^2 + {\omega ^2}}}

Correct Answer

Option B

Solution

Given that, initial angular velocity =

{\omega _0}

and at any instant time t, angular velocity = $\omega$ So when displacement is x then the resultant acceleration f =

\left( {\omega _0^2 - {\omega ^2}} \right)x

So the external force, F =

m\left( {\omega _0^2 - {\omega ^2}} \right)x

............(i) But given that

F \propto \cos \omega t

From (i) we get,

m\left( {\omega _0^2 - {\omega ^2}} \right)x \propto \cos \omega t

.........(ii) From equation of SHM we know,

x = A\sin \left( {\omega t + \phi } \right)

When t = 0 then x = A $\therefore$ A =

A\sin \left( \phi \right)

\Rightarrow A = {\pi \over 2}

$\therefore$

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

Putting value of x in (ii), we get

m\left( {\omega _0^2 - {\omega ^2}} \right)A\cos \omega t \propto \cos \omega t

\Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}

Q93

In a simple harmonic oscillator, at the mean position

A kinetic energy is minimum, potential energy is maximum

B both kinetic and potential energies are maximum

C kinetic energy is maximum, potential energy is minimum

D both kinetic and potential energies are minimum.

Correct Answer

Option C

Solution

K.E = {1 \over 2}k\left( {{A^2} - {x^2}} \right);\,\,\,U = {1 \over 2}k{x^2}

At the mean position

x=0

$\therefore$

K.E. = {1 \over 2}k{A^2} =

Maximum and

U=0

Q94

Two particles

A

and

B

of equal masses are suspended from two massless springs of spring of spring constant

{k_1}

and

{k_2}

, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of

A

and

B

is

A

\sqrt {{{{k_1}} \over {{k_2}}}}

B

{{{{k_2}} \over {{k_1}}}}

C

\sqrt {{{{k_2}} \over {{k_1}}}}

D

{{{{k_1}} \over {{k_2}}}}

Correct Answer

Option C

Solution

Maximum velocity during

SHM

= A\omega = A\sqrt {{k \over m}}

\left[ {\,\,} \right.

$\therefore$

\omega = \sqrt {{k \over m}}

\left. {\,\,} \right]

Here the maximum velocity is same and

m

is also same $\therefore$

{A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}}

$\therefore$

{{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over {{k_1}}}}

Q95

A particle moves with simple harmonic motion in a straight line. In first

\tau s,

after starting from rest it travels a distance

a,

and in next

\tau s

it travels

2a,

in same direction, then:

A amplitude of motion is

3a

B time period of oscillations is

8\tau

C amplitude of motion is

4a

D time period of oscillations is

6\tau

Correct Answer

Option D

Solution

In simple harmonic motion, starting from rest, At

t=0,

x=A

x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

When

t = \tau ,\,\,x = A - a

When

t = 2\,\tau ,\,x = A - 3a

From equation

(i)

A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

As

\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)

From equation

(ii),

(iii)

and

(iv)

{{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1

\Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}

\Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa

\Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a

\Rightarrow {a \over A} = {1 \over 2}

Now,

A-a=A

\cos \omega \tau

\Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}

or,

{{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau

Q96

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

A

{1 \over 2}

B

{3 \over 4}

C

{1 \over 3}

D

{1 \over 4}

Correct Answer

Option B

Solution

K = {1 \over 2}m{\omega ^2}({A^2} - {x^2})

= {1 \over 2}m{\omega ^2}\left( {{A^2} - {{{A^2}} \over 4}} \right)

= {1 \over 2}m{\omega ^2}\left( {{{3{A^2}} \over 4}} \right)

K = {3 \over 4}\left( {{1 \over 2}m{\omega ^2}{A^2}} \right)

Q97

The bob of a simple pendulum executes simple harmonic motion in water with a period

t,

while the period of oscillation of the bob is

{t_0}

in air. Neglecting frictional force of water and given that the density of the bob is

\left( {4/3} \right) \times 1000\,\,kg/{m^3}.

What relationship between

t

and

{t_0}

is true

A

t = 2{t_0}

B

t = {t_0}/2

C

t = {t_0}

D

t = 4{t_0}

Correct Answer

Option A

Solution

t = 2\pi \sqrt {{\ell \over {{g_{eff}}}}} ;\,{t_o}\,\, = 2\pi \sqrt {{\ell \over g}}

m{g_{eff}} = mg - B = my - V \times 100 \times g

$\therefore$

{g_{eff}} = g - {{100} \over {\left( {m/v} \right)}}g

= g - {{1000} \over {{4 \over 3} \times 1000}}g = {g \over 4}

$\therefore$

t = 2\pi \sqrt {{\ell \over {g/4}}} \,\,\,\,\,\,\,\,\,\,\,t = 2{t_0}

Q98

For a body executing S.H.M. : (1) Potential energy is always equal to its K.E. (2) Average potential and kinetic energy over any given time interval are always equal. (3) Sum of the kinetic and potential energy at any point of time is constant. (4) Average K.E. in one time period is equal to average potential energy in one time period. Choose the most appropriate option from the options given below :

A (3) and (4)

B only (3)

C (2) and (3)

D only (2)

Correct Answer

Option A

Solution

In S.H.M. total mechanical energy remains constant and also = =

{{1 \over 4}}

KA2 (for 1 time period)

Q99

An oscillator of mass M is at rest in its equilibrium position in a potential V =

{1 \over 2}

k(x

-

X)2. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5, u = 1, k = 1)

A

{1 \over {\sqrt 3 }}

B

{1 \over 2}

C

{2 \over 3}

D

{3 \over {\sqrt 5 }}

Correct Answer

Option A

Solution

Potential of the given oscillator is

V = {1 \over 2}k{(x - k)^2}

Given: M = 10; m = 5, u = 1; k = 1 Initial momentum of the particle of mass m = mu = m $\times$ 5 = 5m Momentum of (oscillator + particle) after collision = (M + m) Velocity of oscillator after collision = v So, momentum of system = (M + m)v From conservation of linear momentum, we have (M + m) = mu = 5 $\times$ 1 = 5 For second collision, oscillator and particle have momentum in opposite direction.

Net or total momentum is zero.

Likewise after 4th, 6th, 8th, 10th, 12th collision the momentum is zero.

After 12th collision, Mass of oscillator and 12 particles will be (10 + 12 $\times$ 5) = 70 Now, from conservation of linear momentum, for 13th collision, we have

70 \times 0 + 5 \times 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}

Total mass after 13th collision = (10 + 13 $\times$ 5) = 75 Kinetic energy of system

= {1 \over 2}mv{'^2}

\Rightarrow KE = {1 \over 2} \times 75 \times {1 \over {15}} \times {1 \over {15}}

\Rightarrow {1 \over 2}k{A^2} = {1 \over 2} \times {{75} \over {225}} = {1 \over 6}

\Rightarrow {1 \over 2} \times 1 \times {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}

\Rightarrow A = {1 \over {\sqrt 3 }}

Q100

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)

A 4

B 7

C 6

D 5

Correct Answer

Option B

Solution

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency. $\therefore$ (2n + 1) f0 $\le$ 20,000 (f0 is fundamental frequency = 1.5 KHz) $\therefore$ n = 6 $\therefore$ Total number of overtone that can be heared is 7. (0 to 6)