Physics - Simple Harmonic Motion

Question - 1

Two bodies M and N of equal masses are suspended from two separate massless springs of spring constants ${ k }_{ 1 }$ and ${ k }_{ 2 }$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that on N is

• A $\frac { { k }_{ 1 } }{ { k }_{ 2 } }$
• B $\sqrt { \frac { { k }_{ 1 } }{ { k }_{ 2 } } }$
• C $\frac { { k }_{ 2 } }{ { k }_{ 1 } }$
• D $\sqrt { \frac { { k }_{ 2 } }{ { k }_{ 1 } } }$

Question - 2

A body is executing SHM with an amplitude of 0.1 m. Its velocity while passing through the mean position is 3m/s. Its frequency in Hz is

• A $15\pi$
• B $\frac { 15 }{ \pi }$
• C $30\pi$
• D $25\pi$

Question - 3

Two particles are executing SHMs. The equations of their motions are

${ Y }_{ 1 }=10sin\left( \omega t+\frac { \pi }{ 4 } \right)$

${ Y }_{ 2 }=5sin\left( \omega t+\frac { \sqrt { 3 } \pi }{ 4 } \right)$

What is the ratio of their amplitudes.

• A 1:1
• B 2:1
• C 1:2
• D None of these

Question - 4

What is the maximum acceleration of the particle doing the SHM?

$Y=2sin\left[ \frac { \pi t }{ 2 } +\phi \right]$ where Y is in cm

• A $\frac { \pi }{ 2 } cm/{ s }^{ 2 }$
• B $\frac { { \pi }^{ 2 } }{ 2 } cm/{ s }^{ 2 }$
• C $\frac { { \pi } }{ 4 } cm/{ s }^{ 2 }$
• D $\frac { { { \pi }^{ 2 } } }{ 4 } cm/{ s }^{ 2 }$

Question - 5

The amplitude and the time period in a SHM is 0.5 cm and 0.4 s respectively. If the initial phase is $\frac { \pi }{ 2 }$ rad, then the equation of SHM will be

• A $Y=0.5\quad sin\quad 5\pi t$
• B $Y=0.5\quad sin\quad 4\pi t$
• C $Y=0.5\quad sin\quad 2.5\pi t$
• D $Y=0.5\quad cos\quad 5\pi t$

Question - 6

A particle is executing simple harmonic motion with a period of T seconds and amplitude a meter. The shortest time it takes to reach a point $\frac { a }{ \sqrt { 2 } }$ m from its mean position in seconds is

• A T
• B $\frac { T }{ 4 }$
• C $\frac { T }{ 8 }$
• D $\frac { T }{ 16 }$

Question - 7

A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the amplitude

• A $\frac { T }{ 2 }$
• B $\frac { T }{ 4 }$
• C $\frac { T }{ 8 }$
• D $\frac { T }{ 12 }$

Question - 8

A particle executing SHM of amplitude 4 cm and T=4 s.The time taken by it to move from positive extreme position to half the amplitude is

• A 1 s
• B $\frac { 1 }{ 3 }$s
• C $\frac { 2 }{ 3 }$s
• D $\sqrt { \frac { 3 }{ 2 } }$ s

Question - 9

Two simple harmonic motions are represented by the equations ${ y }_{ 1 }=0.1\quad sin\left( 100\pi t+\frac { \pi }{ 3 } \right)$ and ${ y }_{ 2 }=0.1\quad cos\pi t$. The phase difference of velocity of particle 1 with respect to the velocity of particle 2 at time t=0 is

• A $\frac { -\pi }{ 3 }$
• B $\frac { \pi }{ 6 }$
• C $\frac { -\pi }{ 6 }$
• D $\frac { \pi }{ 3 }$

Question - 10

A particle executing simple harmonic motion has an amplitude of 6 cm. Its acceleration at a distance of 2 cm from the mean position is 8 cm /${ s }^{ 2 }$. The maximum speed of the particle is

• A 8 cm/s
• B 12 cm/s
• C 16 cm/s
• D 24 cm/s