A particle moves on the X-axis according to the equation x = A + B sinωt. The motion is simple harmonic with amplitude

1. A

2. B

3. A + B

4. \(\sqrt{A^2+B^2}\)

The figure represents two simple harmonic motions.

The parameter which has different values in the two motions is:
1. amplitude
2. frequency
3. phase
4. maximum velocity

The total mechanical energy of a spring-mass system in simple harmonic motion is \(E=\frac12m~\omega^2A^2\). Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will

1. become 2E

2. become E/2

3. become \(\sqrt E\)

4. remain E

The average energy in one time period in simple harmonic motion is:
1. \(\frac{1}{2} m \omega^{2} A^{2}\) 
2. \(\frac{1}{4} m \omega^{2} A^{2}\) 
3. \(m \omega^{2} A^{2}\)
4. zero

A particle executes simple harmonic motion with a frequency \(\nu.\) The frequency with which the kinetic energy oscillates is:
1. \(\nu/2\)
2. \(\nu\)
3. \(2\nu\)
4. zero

A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the spring is divided in two equal parts and one part is used to continue the simple harmonic motion, the time period will

1. remain T

2. become 2T

3. become T/2

4. become T/\(\sqrt2\)

Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is

1. \(k_{1} / k_{2}\)

2. \(\sqrt{k_{1} / k_{2}}\)

3. \(k_{2} / k_{1}\)

4. \(\sqrt{k_{2} / k_{1}}\)

A spring-mass system oscillates with a frequency \(\nu.\) If it is taken in an elevator slowly accelerating upward, the frequency will:
1. increase
2. decrease
3. remain same
4. become zero

A spring-mass system oscillates in a car. If the car accelerates on a horizontal road, the frequency of oscillation will:
1. increase
2. decrease
3. remain same
4. become zero

A pendulum clock that keeps the correct time on the earth is taken to the moon. It will run:
(Take \(g_{\text{moon}}=\frac{g_{\text{earth}}}{6}\))
1. at the correct rate
2. \(6\) times faster
3. \(\sqrt6\) times faster
4. \(\sqrt6\) times slower