Question #173356

A ball weighing 20g, attached to a spring, performs a harmonic oscillation with a frequency of 10 Hz and an amplitude of A = 10cm.
1. Find the maximum force with which the spring acts on the ball.
2. Find the speed of the ball as it passes through the equilibrium position.
3. Find the mechanical energy of the spring pendulum

Expert's answer

1) Let's first find the spring constant of the spring:


f=12πkm,f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}},k=4π2f2m=4π2(10 Hz)20.02 kg=79 Nm.k=4\pi^2f^2m=4\pi^2\cdot(10\ Hz)^2\cdot0.02\ kg=79\ \dfrac{N}{m}.

Then, we can find the maximum force with which the spring acts on the ball:


Fmax=kA=79 Nm0.1 m=7.9 N.F_{max}=kA=79\ \dfrac{N}{m}\cdot0.1\ m=7.9\ N.

2) The speed of the ball as it passes through the equilibrium position can be found as follows:


vmax=Aω=2πfA=2π10 Hz0.1 m=6.28 ms.v_{max}=A\omega=2\pi fA=2\pi\cdot10\ Hz\cdot0.1\ m=6.28\ \dfrac{m}{s}.

3) We can find the mechanical energy of the spring pendulum as follows:


E=12kA2=1279 Nm(0.1 m)2=0.395 J.E=\dfrac{1}{2}kA^2=\dfrac{1}{2}\cdot79\ \dfrac{N}{m}\cdot(0.1\ m)^2=0.395\ J.
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