Question

With a 50 g mass at its end, a spring undergoes SHM with a frequency of 0.7 Hz. How much work is done in stretching the spring 15 cm from its unstretched length? How much energy is then stored in the spring?

Answer 1

a) Let's first find the spring constant from the definition of frequency of SHM:

$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}},$$ $$k=4\pi^2f^2m,$$ $$k=4\pi^2\cdot(0.7\ Hz)^2\cdot0.05\ kg=0.97\ \dfrac{N}{m}.$$

Then, we can find the work done in stretching the spring as follows:

$$W=\dfrac{1}{2}k(x_f^2-x_i^2),$$ $$W=\dfrac{1}{2}\cdot0.97\ \dfrac{N}{m}\cdot((0.15\ m)^2-0)=0.011\ J.$$

b) The work done in stretching the spring is equal to the elastic potential energy stored in the spring:

$$W=PE_s=0.011\ J.$$