Question

A particle with mass 1.23 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.847 m and a duration of 125 s for 69 cycles of oscillation. Find the frequency, ?, the speed at the equilibrium position, ?max, the spring constant, ?, the potential energy at an endpoint, ?max, the potential energy when the particle is located 41.3% of the amplitude away from the equiliibrium position, ?, and the kinetic energy, ?, and the speed, ?, at the same position.

?= Hz ?max= m/s

?= N/m

?max= J

?= J

?= J ?= m/s

Answer 1

Potential energy is maximum when the extension out comprehension in the spring is maximum extension on comprehnesion is

x = A = 0.847

Vmax = 1/2 Kx²

1/2 K (A)²

= 1/2 (11.22) (0.847)

Vmax = 4.75 J

When the particle is located 41.3% of the amplitude than the extension in the spring is

x = A X 41.3/100 = 0.847 X 41.3/100

x = 0.349

than V = 1/2 Kx² = 1/2 X 11.22 X (0.349)²

V = 0.683 J

for energy conservation

Vmax = 0 + K

4.75 = 0.683 + K

K = 4.067 J

than K = 1/2 mv²

4.067 = 1/2 X 1.23 X theta²

V² = 3.452

V = 1.857 m/s