Oscillations · Question 15

Identify the function which represents a non-periodic motion.

1. ( t + /4)
2. e- t
3. t
4. t + t

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is :

1. 4n
2. n
3. 2n
4. 3n

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is -

1. 0.628 s
2. 0.0628 s
3. 6.28 s
4. 3.14 s

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :

1. 3 2 rad
2. 2 rad
3. zero
4. rad

The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the figure. [figure] y- projection of the radius vector of rotating particle P is :

1. y(t) = 4 ( t2 ), where y in m
2. y(t) = 3 ( 3 t2 ), where y in m
3. y(t) = 3 ( t2 ), where y in m
4. y(t) = -3 2 t, where y in m

Average velocity of a particle executing SHM in one complete vibration is :

1. A
2. A 22
3. A 2
4. zero

The displacement of a particle executing simple harmonic motion is given by y = A0 + A t + B t. Then the amplitude of its oscillation is given by :

1. A2 + B2
2. A + B
3. A + A2 + B2
4. A02 + (A + B)2

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s-2 at a distance of 5 m from the mean position. The time period of oscillation is

1. 2 s
2. s
3. 2 s
4. 1 s

A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is

1. 1 : 9
2. 1 : 11
3. 1 : 14
4. 1 : 6

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

1. 52
2. 4 5
3. 2 3
4. 5