A mass M is suspended from a spring and oscillates with a period of 0.940 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.

The time, t = 7.99seconds

Explanation:

Total energy of SHM system is:

E = ½kA²

0.50E = 0.50 (½kAo²) = ½kA²

A = Ao/√2 = 0.707Ao

The amplitude after each oscillation is given by:

A (n) = Ao (0.96) ^n

Solve for n by setting this equal to half-energy amplitude:

Ao (0.96) ^n = Ao/√2

ln (Ao (0.96) ^n) = ln (Ao/√2)

ln (Ao) + n*ln (0.96) = ln (Ao) - ln (√2)

n = - ln (√2) / ln (0.96)

n = 8.5

Find time from n & period T:

Period, T = 0.940seconds

t = 8.5 * 0.940

t = nT = 7.99s