Simple Harmonic Motion — Page 5 | JEE Physics

Q41

The time period of a simple pendulum is given by

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

. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-

A 2%

B 3%

C 5%

D 4%

Correct Answer

Option B

Solution

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

{T^2} = 2\pi \left( {{l \over g}} \right)

g = 2\pi {l \over {{T^2}}}

{{\Delta g} \over g} = {{\Delta l} \over l} + {{2\Delta T} \over T}

{{\Delta g} \over g} = {{1 \times {{10}^{ - 3}}} \over {1 \times {{10}^{ - 2}}}} + {{2 \times 1} \over {100}}

{{\Delta g} \over g} = 0.02 + 0.01 = 0.03

100 \times {{\Delta g} \over g} = 0.03 \times 100 = 3\%

{{\Delta g} \over g} \times 100 = 3\%

Q42

The function of time representing a simple harmonic motion with a period of

{\pi \over \omega }

is :

A cos(

\omega

t) + cos(2

\omega

t) + cos(3

\omega

t)

B sin2(

\omega

t)

C sin(

\omega

t) + cos(

\omega

t)

D 3cos

\left( {{\pi \over 4} - 2\omega t} \right)

Correct Answer

Option D

Solution

General equation of SHM x = A sin( $\omega$ 't $\pm$ $\phi$ ) We know, $\omega$ =

{{2\pi } \over T}

Given,

T = {\pi \over \omega }

$\therefore$ $\omega$ ' =

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

= 2 $\omega$ $\therefore$ Equation becomes, x = a sin(2 $\omega$ t $\pm$ $\phi$ ) Here coefficient of t is 2 $\omega$ . you can see only option (D) has coefficient 2 $\omega$ .

Q43

A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as :

A

T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}}

B

T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}}

C

T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}}

D

T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}}

Correct Answer

Option D

Solution

{v^2} = {\omega ^2}({A^2} - {x^2})

{A^2} = x_1^2 + {{v_1^2} \over {{\omega ^2}}} = x_2^2 + {{v_2^2} \over {{\omega ^2}}}

{\omega ^2} = {{v_2^2 - v_1^2} \over {x_1^2 - x_2^2}}

T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}}

Q44

T0 is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to

{1 \over {16}}

times of its initial value, the modified time period is :

A 4 T0

B

{1 \over {4}}

T0

C T0

D 8

\pi

T0

Correct Answer

Option B

Solution

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

T' = {{{T_0}} \over 4}

Q45

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is

{{3E} \over 4}

then its displacement 'y' is given by :

A y = a

B

y = {a \over {\sqrt 2 }}

C

y = {{a\sqrt 3 } \over 2}

D

y = {a \over 2}

Correct Answer

Option D

Solution

E = {1 \over 2}K{a^2}

{{3E} \over 4} = {1 \over 2}K({a^2} - {y^2})

$\Rightarrow$

{3 \over 4} \times {1 \over 2}K{a^2} = {1 \over 2}K({a^2} - {y^2})

$\Rightarrow$

{y^2} = {a^2} - {{3{a^2}} \over 4}

$\Rightarrow$

y = {a \over 2}

Q46

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density

{1 \over 4}

times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-

A T

B

{3 \over 2}

T

C

{3 \over 4}

T

D

{4 \over 3}

T

Correct Answer

Option D

Solution

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

When bob is immersed in liquid mgeff = mg $-$ Buoyant force mgeff = mg $-$ v $\sigma$ g ( $\sigma$ = density of liquid)

= mg - v{\rho \over 4}g

= mg - {{mg} \over 4} = {{3mg} \over 4}

$\therefore$

{g_{eff}} = {{3g} \over 4}

{T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}}

{l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}

By solving

{T_1} = {4 \over 3}2\pi \sqrt {l/g}

{T_1} = {{4T} \over 3}

Q47

The equation of a particle executing simple harmonic motion is given by

x = \sin \pi \left( {t + {1 \over 3}} \right)m

. At t = 1s, the speed of particle will be (Given :

\pi

= 3.14)

A 0 cm s

-

1

B 157 cm s

-

1

C 272 cm s

-

1

D 314 cm s

-

1

Correct Answer

Option B

Solution

x = \sin \left( {\pi t + {\pi \over 3}} \right)m

\Rightarrow {{dx} \over {dt}} = \pi \cos \left( {\pi t + {\pi \over 3}} \right)

= \pi \cos \left( {\pi + {\pi \over 3}} \right)

at

t = 1\,s

= - {\pi \over 2}

m/s or

\left| {{{dx} \over {dt}}} \right| = 157

cm/s

Q48

The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :

A 6 s

B 8 s

C 12 s

D 36 s

Correct Answer

Option D

Solution

Time taken by the harmonic oscillator to move from mean position to half of amplitude is

{T \over {12}}

So,

{T \over {12}}

= 3 T = 36 sec.

Q49

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with

{g \over 6}

vertically upwards then the time period will be : (Where g = acceleration due to gravity)

A

\sqrt {{6 \over 5}} T

B

\sqrt {{5 \over 6}} T

C

\sqrt {{6 \over 7}} T

D

\sqrt {{7 \over 6}} T

Correct Answer

Option C

Solution

$T^{\prime}=2 \pi \sqrt{\dfrac{I}{g_{\text{eff }}}}$ $T^{\prime}=2 \pi \sqrt{\dfrac{I}{g+\dfrac{g}{6}}}=2 \pi \sqrt{\dfrac{6 l}{7 g}}$ $\Rightarrow T^{\prime}=\sqrt{\dfrac{6}{7}} T$

Q50

Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :

A 1 : 2

B 3 : 2

C 3 : 1

D 2 : 3

Correct Answer

Option B

Solution

{\omega _1}{A_1} = {\omega _2}{A_2}

\Rightarrow {{{A_1}} \over {{A_2}}} = {{{\omega _2}} \over {{\omega _1}}}

= \sqrt {{{{k_2}} \over {{m_2}}}} \times \sqrt {{{{m_1}} \over {{k_1}}}} = \sqrt {{{9k} \over {100}} \times {{50} \over {2k}}} = {3 \over 2}