A wave pulse travels down a slinky The mass of the slinky is

A wave pulse travels down a slinky. The mass of the slinky is m = 0.89 kg and is initially stretched to a length L = 6.7 m. The wave pulse has an amplitude of A = 0.29 m and takes t = 0.418 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.46 Hz.

Now the slinky is stretched to twice its length (but the total mass does not change). What is the new tension in the slinky? (assume the slinky acts as a spring that obeys Hooke’s Law).

Solution

double the distance new velocity = 2 L / t

= 2 * 6.7 / 0.418

= 32.06

Tension = mass * 32.06² / 2 L

= 0.89 * 32.06² / 2 * 6.7

= 68.27 N