Question

simple harmonic motion
when a mass on a spring is at maximum displacement, what quantity is at a minimum?
net force
acceleration
amplitude
velocity

Explanation:

1. Recall the concepts of simple harmonic motion (SHM) for a mass - spring system:

• The force exerted by a spring is given by Hooke's law, \(F = -kx\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. The net force on the mass is equal to the spring force (neglecting other forces like air resistance). When the displacement \(x\) is maximum (at the amplitude), the magnitude of the net force \(F = k|x|\) is also maximum (since \(|x|\) is maximum).
• According to Newton's second law \(F = ma\), so acceleration \(a=\frac{F}{m}=\frac{-kx}{m}\). When \(|x|\) is maximum, the magnitude of the acceleration \(|a|=\frac{k|x|}{m}\) is also maximum.
• The amplitude of the motion is the maximum displacement from the equilibrium position. So, at maximum displacement, the amplitude is just equal to that maximum displacement; it does not change in value (the amplitude is a constant for a given SHM, assuming no energy loss).
• The velocity of the mass in SHM is given by \(v=\pm\omega\sqrt{A^{2}-x^{2}}\), where \(\omega=\sqrt{\frac{k}{m}}\) is the angular frequency and \(A\) is the amplitude. When \(x = A\) (maximum displacement), \(v=\pm\omega\sqrt{A^{2}-A^{2}} = 0\). So, at maximum displacement, the velocity is at its minimum (zero) value.

Answer:

velocity