If a uniform spring is cut in half, what is the force constant of each half? Justify your answer. How would the frequency of SHM using a half-spring differ from the frequency using the same mass and the entire spring?

A simple harmonic motion is characterized by a trajectory that can be described by the trigonometric functions, sine, and cosine. There are a lot of motions that can be described by simple harmonic motion, such as the motion of a swinging simple pendulum.

A uniform spring can be considered as two identical springs which are connected in series.

When a force F is applied to it and the total extension is x the from Hooke's law$\mathrm{F}=‐\mathrm{kx}$, where k is the force constantof the entire spring.

Since the spring is uniform, each half of the spring stretches by equal amount, which is x/2.

Hence the force constant of each spring is${\mathrm{k}}^{\mathrm{\varphi }}=‐\frac{\mathrm{F}}{\mathrm{x}/2}=‐2\frac{\mathrm{F}}{\mathrm{x}}=2\mathrm{k}$.

Hence the force constant ofeach spring is twice that of the fulllength spring.

The frequency of the SHM is given by $\mathrm{\omega }=\sqrt{\mathrm{k}/\mathrm{m}}$.Since the force constant increases by 2 times, the frequency of simple harmonic motion increases by $\sqrt{2}$times of the frequency of the entire spring when the same is used.