A particle with mass 1.73 kg oscillates horizontally at the end of a horizontal

Question

A particle with mass 1.73 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.845 m and a duration of 129 s for 69 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 49.3% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position. need help on this

Explanation / Answer

1)Frequency=69/129=0.534 hz

2)f=1/(2?)*sqrt(k/m)

k=19.47 N/m

3)The speed at eq. position be v.

U=0.5kA2=0.5mv2

v=2.83m/s

4)Potential energy at endpoint=0.5kx2=6.95J

5) U1=0.5*19.47*(0.493*0.845)2=1.689J

6)K= Total energy-U1=6.95J-1.689J=5.261J

7) Speed=sqrt(2K/m)=4.744m/s

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