Simple Harmonic Motion — Page 8 | NEET Physics

Q71

The period of oscillation of a simple pendulum is

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

. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

A 1 %

B 5 %

C 2 %

D 3 %

Correct Answer

Option D

Solution

Given

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

\Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}

\Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}

[ as

T = {t \over n}

] So, percentage error in

g

=

{{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta t} \over t} \times 100

=

{{0.1} \over {20.0}} \times 100 + 2 \times {1 \over {90}} \times 100

= 2.72 % = 3 %

Q72

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant

t = {T \over 4}s

starting from mean position. Assume that the initial phase of the oscillation is zero.

A 0.62 J

B 6.2

\times

10

-

3 J

C 1.2

\times

103 J

D 6.2

\times

103 J

Correct Answer

Option A

Solution

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

0.2 = 2\pi \sqrt {{{0.5} \over k}}

k = 50 $\pi$ 2 $\approx$ 500 x = A sin ( $\omega$ t + $\phi$ ) = 5 cm sin

\left( {{{\omega T} \over 4} + 0} \right)

= 5 cm sin

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

= 5 cm

PE = {1 \over 2}k{x^2}

= {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}

= 0.6255

Q73

In a linear Simple Harmonic Motion (SHM) (A) Restoring force is directly proportional to the displacement. (B) The acceleration and displacement are opposite in direction. (C) The velocity is maximum at mean position. (D) The acceleration is minimum at extreme points. Choose the correct answer from the options given below:

A

{\text{(A), (B) and (D) only }}

B (C) and (D) only

C (A), (B) and (C) Only

D (A), (C) and (D) only

Correct Answer

Option C

Solution

The correct options are: (A) Restoring force is directly proportional to the displacement. - True (this is a defining characteristic of SHM) (B) The acceleration and displacement are opposite in direction. - True (the acceleration is proportional to the displacement but in the opposite direction) (C) The velocity is maximum at mean position. - True (the velocity is zero at the extreme positions and reaches a maximum at the mean position) (D) The acceleration is minimum at extreme points. - False (the acceleration is maximum at the extreme points and zero at the mean position)

Q74

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :

A 0.77

B 0.57

C 0.37

D 0.17

Correct Answer

Option C

Solution

Initially : After putting 2 masses of each 'm' at a distance

{L \over 2}

from center : We know, Time period (T) = 2 $\pi$

\sqrt {{{\rm I} \over C}}

$\therefore$ T $\propto$

\sqrt {\rm I}

$\therefore$ Frequency (f) $\propto$

\sqrt {{1 \over {\rm I}}}

$\therefore$

{{{f_1}} \over {{f_2}}}

=

\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}

Also given that, After putting two masses 'm' at both end new frequency becomes 80% of initial frequency. $\therefore$ f2 = 0.8f1 $\therefore$

{{{f_1}} \over {0.8{f_1}}}

=

\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}}

$\therefore$

{{{{{\rm I}_1}} \over {{{\rm I}_2}}}}

= 0.64 Initial moment of inertia of the system,

{{{\rm I}_1}}

=

{{M{{\left( {2L} \right)}^2}} \over {12}}

Final moment of inertia of the system, I2 =

{{M{{\left( {2L} \right)}^2}} \over {12}}

+ 2

\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)

$\therefore$

{{M{{\left( {2L} \right)}^2}} \over {12}}

= 0.64

\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]

$\Rightarrow$

{{M{L^2}} \over {3 \times 0.64}}

=

{{M{L^2}} \over 3}

+

{{M{L^2}} \over 2}

$\Rightarrow$

{M \over {1.92}} - {M \over 3} = {m \over 2}

$\Rightarrow$

{{1.08M} \over {3 \times 1.92}}

=

{m \over 2}

$\Rightarrow$

{m \over M}

=

{{1.08 \times 2} \over {3 \times 1.92}}

= 0.37

Q75

A mass

M

is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes

SHM

of time period

T.

If the mass is increased by

m.

the time period becomes

{{5T} \over 3}

. Then the ratio of

{{m} \over M}

is

A

{3 \over 5}

B

{25 \over 9}

C

{16 \over 9}

D

{5 \over 3}

Correct Answer

Option C

Solution

The time period of a simple harmonic motion (SHM) performed by a mass-spring system is given by the formula:

T = 2\pi \sqrt{{M \over k}}

where: T is the time period, M is the mass of the object, and k is the spring constant.

We know that if the mass is increased by m, the time period becomes $\dfrac{5T}{3}$ .

We can set up an equation for this new scenario:

\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}

Since we know that $T = 2\pi \sqrt{\dfrac{M}{k}}$ , we can substitute T in the equation above:

\frac{5}{3} \cdot 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M + m}{k}}

Squaring both sides of the equation to eliminate the square root, we get:

\frac{25}{9} \cdot \frac{M}{k} = \frac{M + m}{k}

Solving for $\dfrac{m}{M}$ , we get:

\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}

Q76

In forced oscillation of a particle the amplitude is maximum for a frequency

{\omega _1}

of the force while the energy is maximum for a frequency

{\omega _2}

of the force; then

A

{\omega _1} < {\omega _2}

when damping is small and

{\omega _1} > {\omega _2}

when damping is large

B

{\omega _1} > {\omega _2}

C

{\omega _1} = {\omega _2}

D

{\omega _1} < {\omega _2}

Correct Answer

Option C

Solution

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition) $\therefore$

{\omega _1} = {\omega _2}

Q77

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%.

Q78

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)}}

Q79

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

Q80

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}}}}