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A pendulum of length L is undergoing simple harmonic motion. When the length of the string is halved, how does the period of the pendulum change?
Answer:
The period of the pendulum decreases by a factor of √2 since,
T=2π*√(L/g)
becomes T=2π*√(L/2g). Therefore the period would become a smaller number specifically by a factor of √2
The axis that's along the length of the string.
Since there is no motion along the radial axis, it must be true that the tension in the string and the weight are equal.
T=mgcosƟ
Angular Frequency
ω = √(g/L)
Period= 2π/ω= 2π*√(L/g)
Frequency of Oscillation
f=ω/2π= (1/2π)(√(g/L))
where, L represents the length of the string, and g represents the acceleration of gravity.
The tangential weight of the mass provides the restoring force that allows the mass to swing towards the equilibrium position.
Fnet,t =-mgsinƟ=ma
rearranging, for acceleration leads us to,
a=-gsinƟ
From Physics for Scientists and Engineers: An Interactive Approach
The forces acting on the pendulum is the weight itself mg, and the tension of the string.
A simple pendulum consists of a mass attached to a string of negligible mass. The other end of the string is attached to a frictionless support.
When the mass is displaced from its vertical equilibrium position and released, the pendulum would swing in the path as shown in the dotted red line.