A particle of mass 200 g executes SHM. The restoring force is provided by a spring of force constant 80 N m-1. The time period of oscillation is
Question
A particle of mass 200 g executes SHM. The restoring force is provided by a spring of force constant 80 N m-1. The time period of oscillation is
Solution
The time period of oscillation for a particle executing Simple Harmonic Motion (SHM) is given by the formula:
T = 2π √(m/k)
where:
- T is the time period,
- m is the mass of the particle, and
- k is the spring constant.
Given in the problem:
- m = 200 g = 0.2 kg (since 1 kg = 1000 g)
- k = 80 N/m
Substituting these values into the formula, we get:
T = 2π √(0.2/80)
Solving this will give us the time period of oscillation.
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