Question

1. A 0.45 kg object mass attached to a spring whose spring constant is 600 n/m executes simple harmonic motion. its maximum speed is 3.0 m/s the maximum acceleration is:

Answer 1

Suppose the equation of position is: x = Asin(wt)

where

• A is amplitude
• w = √(k/m) is angular frequency, where k is spring Constant and m is mass attached to spring
• t is time

w = √(k/m) = √(600 N/m / 0.45 kg) = 36.515 s^{​​​​​​-1}

V = dx/dt = Awcos(wt)

Maximum speed will be when cos(wt) = 1

V_{​​​​​max} = Aw = 36.515A m/s = 3.0 m/s

=> A = 0.082 m

a = dV/dt = -Aw²sin(wt)

a_{​​​​​​Max} = Aw² = (0.082 m)*(36.515)² = 109.54 m/s²

Hence, Maximum acceleration is 109.54 m/s²