Question

A 0.50 kg mass vibrates according to the equation x=0.50cos6.80t, where x is in meters and t is in seconds.

A) Determine the amplitude.

B) Determine the frequency.

C) Determine the total energy

D) Determine the kinetic energy when x = 0.36 m .

E) Determine the potential energy when x = 0.36 m .

Answer 1

Given

m = 0.50 kg.

x = 0.50*cos*6.80*t

we know

x = A*cos*2*f*t   

(a)

The amplitude must be 0.50 m

b)

For Frequency

f = / 2

f = 6.80rad/s / 2

f = 1.08 Hz

(c)   

Total energy

TME = max KE = ½*m*(Vmax)² = ½*m*(A)²

TME = ½ * 0.50kg * (0.50m * 6.80rad/s)²

TME = 2.89 J

(d) & (e)

when x = 0.36 m

k = m*²

k = 0.50kg * (6.80 rad/s)²

k = 23.12 kg/s²

PE = ½ * k * x²

PE = ½ * 23.12 N/m * (0.36m)²

PE = 1.498 J

PE = 1.5 J

KE = TME - PE

KE = 2.89 – 1.498

KE = 1.392 J

KE = 1.4 J