Answer the following questions : (a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: T = 2π √(m/k) A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum? (b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π √( l /g) Think of a qualitative argument to appreciate this result. (c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ? (d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ?
a)
Time period of a particle is given as,
T = 2 π m k .
Where, the force constant is k and the mass is m .
Since, in the case of spring, the force constant does not depend on mass and in the case of simple pendulum the force constant is directly proportional to mass so, it can be written as,
m k = const
Thus, the time period of a pendulum is independent of mass of the pendulum.
b)
The time period is given as,
T = 2 π l g
The restoring force for the bob of the pendulum is given as,
F = − m g sin θ
For small angle θ , sin θ ≈ θ .
But for large angle θ , sin θ ≠ θ . This means the effective value of g decreases.
Now, the time period of the pendulum will be,
T ' = 2 π l g '
As g ' < g therefore, the time period of the pendulum T ' > T .
Thus, for the large angles of oscillation is greater than 2 π l g .
c)
The electronic system and spring system is used in the wrist watch to give the time and it will be independent from acceleration due to gravity.
Thus, the wrist watch will give the correct time because it is independent from acceleration due to gravity.
d)
It can be written in the case of free fall of cabin the value of acceleration due to gravity will be zero and then the frequency of oscillation will also be zero.
Thus, the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity is zero.
a)
Time period of a particle is given as,
Where, the force constant is
Since, in the case of spring, the force constant does not depend on mass and in the case of simple pendulum the force constant is directly proportional to mass so, it can be written as,
Thus, the time period of a pendulum is independent of mass of the pendulum.
b)
The time period is given as,
The restoring force for the bob of the pendulum is given as,
For small angle
But for large angle
Now, the time period of the pendulum will be,
As
Thus, for the large angles of oscillation is greater than
c)
The electronic system and spring system is used in the wrist watch to give the time and it will be independent from acceleration due to gravity.
Thus, the wrist watch will give the correct time because it is independent from acceleration due to gravity.
d)
It can be written in the case of free fall of cabin the value of acceleration due to gravity will be zero and then the frequency of oscillation will also be zero.
Thus, the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity is zero.
.
A simple pendulum executes SHM approximately. Why then is the time
period of a pendulum independent of the mass of the pendulum?
.
Think of a qualitative argument to appreciate this result.