Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:
. 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. 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 cabinthat is freely falling under gravity?
Answers
(a) For simple pendulum , K is proportional to mass m of particle . hence, m/K becomes constant and doesn't effect the time period .
(b) for small angle sin∅ ≈∅ but for large angle sin∅ < ∅ . since this factor is multiplied to the restoring force ' mgsin∅' is replaced by ' mg∅' . which means effective reduction in g for large angle .
Hence, an increase in time period T over that given by
T = 2π√{L/g}
compare to the case which it is assumed sin∅ = ∅
(c) yes , actually the motion of hands of a wristwatch to indicate time depends on action of the spring and has nothing to do with acceleration due to gravity . you can say that wristwatch is independent of acceleration due to gravity .
(d) in free fall body effective g = 0 . that is gravity is disapear . ( I mean weightless)
So, T = 2π√( L/g)
= 2π√(L/0) = ∞
Hence, frequency = 1/T = 1/∞ = 0
Answer:
Explanation:(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.
(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:
F=–mgsinθ
where,
F= Restoring force
m= Mass of the bob
g= Acceleration due to gravity
θ= Angle of displacement
For small θ,sinθ≃θ
For large θ, sinθ is greater than θ.
This decreases the effective value of g.
Hence, the time period increases as:
T=2π √l/g
(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
(d) Gravity disappears for a object under free fall, so frequency is zero.