Question

A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The wave pulse has an amplitude of A = 0.23 m and takes t = 0.478 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.48 Hz.

1) What is the speed of the wave pulse? m/s2) What is the tension in the slinky?N3) What is the wavelength of the wave pulse?mPLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS

Answer 1

1.
`v=14.2259\ m.s^(-1)`

2.
`F_T=25.8924\ N`

3.
`\lambda=29.6373\ m`

Step-by-step explanation:

Given:

• mass of slinky,
  `m=0.87\ kg`

• length of slinky,
  `L=6.8\ m`

• amplitude of wave pulse,
  `A=0.23\ m`

• time taken by the wave pulse to travel down the length,
  `t=0.478\ s`

• frequency of wave pulse,
  `f=0.48\ Hz=0.48\ s^(-1)`

1.

`\rm Speed\ of\ wave\ pulse=Length\ of\ slinky/ time\ taken\ by\ the\ wave\ to\ travel`

`v=(6.8)/(0.478)`

`v=14.2259\ m.s^(-1)`

2.

Now, we find the linear mass density of the slinky.

`\mu=(m)/(L)`

`\mu=(0.87)/(6.8)\ kg.m^(-1)`

We have the relation involving the tension force as:

`v=\sqrt{(F_T)/(\mu) }`

`14.2259=\sqrt{(F_T)/((0.87)/(6.8)) }`

`202.3774=F_T* (6.8)/(0.87)`

`F_T=25.8924\ N`

3.

We have the relation for wavelength as:

`\lambda=(v)/(f)`

`\lambda=(14.2259)/(0.48)`

`\lambda=29.6373\ m`