The position of a 54 g oscillating mass is given by xt24cmco

The position of a 54 g oscillating mass is given by x(t)=(2.4cm)cos13t, where t is in seconds. Determine the total energy of the oscillator.

Solution

Amplitude = 2.4 cm = 2.4*10^-2 m

period T = 2*3.14/13 = 0.483 sec

T = 2*pi*(m/k)^0.5

k = 0.054*(2*3.14/0.483)^2 = 9.128 N/m

total energy = 0.5*k*A^2 = 0.5*9.128*0.024^2 = 2.62*10^-3 J

Total energy = 2.62*10^-3 J

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