Question

A rope is vibrating in such a manner that three equal-length segments are found to be vibrating up and down with 321 complete cycles in 20.0 seconds. Waves travel at speeds of 26.4 m/s in the rope. What is the length of the rope?

Answer 1

2.48 m is the length of the rope.

Step-by-step explanation:

Cycle period is 321

Time is 20 second

Wave travels at speeds = 26.4 m/s

We know that,

`\text { Frequency }=(1)/(T) \text { (For one cycle) }`

Frequency required for 321 complete cycle in 20 seconds is

`\text { Frequency }(f)=(321)/(20)`

Frequency = 16.05 hz

We know that,

`\text { Wavelength }=\frac{\text { wave velocity }}{\text { frequency }}`

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

λ = wavelength, the distance between "wave crests" (m)

v = wave velocity, the "speed" that waves are moving in a direction (m/s)

f = frequency, (cycles/ or Hz)

`\lambda=(26,4)/(16.05)`

λ = 1.65 m

As per given question "length of the rope has three equal length segment"

`\mathrm{Length of the rope}=(3)/(2) \lambda`

`\mathrm{Length of the rope }=(3)/(2) * 1.65`

`\mathrm{Length of the rope}=1.5 * 1.65`

Length of the rope = 2.48 m

Therefore, length of the rope is 2.48 m.