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

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