Question

An undamped 1.23 kg horizontal spring oscillator has a spring constant of 37.4 N/m. While oscillating, it is found to have a speed of 2.48 m/s as it passes through its equilibrium position. What is its amplitude of oscillation?

Answer 1

0.45 m

Step-by-step explanation:

m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s

velocity is maximum when it passes through the mean position.

`T = 2\pi \sqrt{(m)/(k)}`

`T = 2\pi \sqrt{(1.23)/(37.4)}`

T = 1.139 sec

w = 2 π / T

w = 2 x 3.14 / 1.139

w = 5.51 rad / s

Vmax = w A

Where, A be the amplitude

2.48 = 5.51 A

A = 2.48 / 5.51 = 0.45 m