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

Answer:

2.48 m is the length of the rope.

Explanation:

Cycle period is 321

Time is 20 second

Wave travels at speeds = 26.4 m/s

We know that,

$\text { Frequency }=\frac{1}{T} \text { (For one cycle) }$

Frequency required for 321 complete cycle in 20 seconds is

$\text { Frequency }(f)=\frac{321}{20}$

Frequency = 16.05 hz

We know that,

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

$\lambda=\frac{v}{f}$

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)

$\lambda=\frac{26,4}{16.05}$

λ = 1.65 m

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

$\mathrm{Length of the rope}=\frac{3}{2} \lambda$

$\mathrm{Length of the rope }=\frac{3}{2} \times 1.65$

$\mathrm{Length of the rope}=1.5 \times 1.65$

Length of the rope = 2.48 m

Therefore, length of the rope is 2.48 m.