# Topic: Oscillations (Test 1)

Topic: Oscillations
Q.1
The pendulum of a wall clock executes
A. an oscillatory motion
B. a translatory motion
C. a rotary motion
D. a circular motion
Answer : Option A
Explaination / Solution:

It executes simple harmonic motion because in this motion restoring force is directly proportional to the displacement towards the mean position at any time.

Workspace
Report
Q.2
for any periodic motion with a period T and displacement = f(t)
A. f (t) = f (t+T )
B. f (t) = f (t+2T )
C. f (t) = f (t+5T )
D. f (t) = f (t+3T )
Answer : Option A
Explaination / Solution:
No Explaination.

Workspace
Report
Q.3
The time period T of a spring system of mass m and spring constant k is given by
A. T=πkm
B. T=mk
C.
D. T=2πmk
Answer : Option D
Explaination / Solution:

If the restoring force of a vibrating or oscillatory system is proportional to the displacement
of the body from its equilibrium position and is directed opposite to the direction of displacement,
the motion of the system is simple harmonic and it is given by

where A, the maximum value of the displacement, is called the amplitude of the motion. If
T is the time for one complete oscillation, then
x(t + T) = x(t)

or
As angular frequency   is given by  and

Then,time period of oscillation of mass m is given by

Workspace
Report
Q.4
In simple harmonic motion the damping force is
A. Inversely proportional to velocity
B. Directly proportional to acceleration
C. Directly proportional to velocity
D. Inversely proportional to acceleration
Answer : Option C
Explaination / Solution:

In real oscillators, friction, or damping, slows the motion of the system. Due to frictional force, the velocity decreases in proportion to the acting frictional force.

where the c is the damping constant

Workspace
Report
Q.5
What is constant in simple harmonic motion?
A. Potential energy
B. Time period
C. Restoring force
D. Kinetic motion
Answer : Option B
Explaination / Solution:

The time period of the SHM is given by  where 'm' be the mass of the body (constant) , 'k' restoring force constant as T depends on 'm' and 'k' and they are constant for the system, so the corresponding Time period of the motion is Constant.

Workspace
Report
Q.6
The kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. Then, the displacement of the body is x percent of the amplitude, where x is
A. 50
B. 67
C. 87
D. 33.0
Answer : Option D
Explaination / Solution:
No Explaination.

Workspace
Report
Q.7
A spring has a certain mass suspended from it and its period for vertical oscillations is  . The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now  .The ratio of  is
A. 2
C.

D.
2
Answer : Option D
Explaination / Solution:
No Explaination.

Workspace
Report
Q.8
A rubber ball with water, having a small hole in its bottom is used as the bob of a simple pendulum. The time-period of such a pendulum:
A. Is a constant
B. First increases and then decreases finally having same value as at the beginning
C. Increases with time
D. Decreases with time
Answer : Option B
Explaination / Solution:
No Explaination.

Workspace
Report
Q.9
Two pendulums have time periods T and 5T/4. They are in phase at their mean positions at some instant of time. What will be their phase difference when the bigger pendulum completes one oscillation?
A. 30
B. 60
C. 45
D. 90
Answer : Option D
Explaination / Solution:

Assuming bigger pendulum as the pendulum of bigger time period i.e.having time period

As the two pendulums start at the same time in simple harmonic motion from the mean position,their initial phase angles are zero

The bigger pendulum completes its one oscillation in time  and comes back to the mean position.During this time the point on reference circle associated with SHM rotates by an angle 2π

If ω represents the angular velocity of the reference point associated with SHM of first pendulum of time period T,then the phase of the this pendulum after  s willbe

Hence the phase difference

= 90

Workspace
Report
Q.10
The length of a second’s pendulum decreases by 0.1percent, then the clock
A. Gains 43.2 seconds per day
B. Loses 43.2 seconds per day
C. Loses 13.5 seconds per day
D. . Loses 7 seconds per day
Answer : Option A
Explaination / Solution:
No Explaination.

Workspace
Report