Assume the acceleration due to gravity is 9.8m/s2.?

Question

1.An aluminum sphere of mass 1.5 kg is attached to a frictionless pivot using mass less string of length 1m.

(a). Assuming the system as a simple pendulum, calculate its period.

(b). The length of the pendulum is adjusted such a way that its period is exactly

one second. Find the length of the pendulum.

(c). A spring mass system with an identical mass oscillates with the same period (i.e. 1s). Find its spring constant.

Answer 1

Formula for a pendulum is T = 2pi*√(l/g)

a) T = 2pi *√(1/9.8) = 2 seconds.

b) T²/4pi²= l/g

     1/4pi² = l/9.8  Cross multiply, l = 0.25 m.

c) T = 2pi*√(m/k)

     T²/4pi² = 1.5/k  cross multiply, k = 37.7.