1. If a mass of 130 grams is attached to the spring, what would be
its elongation? Use the value of K obtained from your
measurements.
2. If two identical springs [each with a spring constant K] were put
together in parallel, what would be their “effective” or resultant
spring constant?
3. If the spring constant, K, is doubled, how would this affect your
answer in question 1?
Answers
(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 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.