Solving Part 6(c): Speed of Transverse Waves on String

[Biosyn]

Homework Statement

Problem is part 6 (c) on page 9 of pdf file.

lodischool.tripod.com/dovesol/DOVE98SOL.pdf

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Homework Equations

λ₈= 2L/8

f₈= 8f₁

V = √(F/(m/L))

The Attempt at a Solution

Okay, I found that the speed of transverse waves is 288m/s
which gives me a tension force of 8.2944N on the string.

λ₄= (2L)/4
λ₄ = 0.6m

f₁λ₄=v
(4 x 120hz)(0.6m) = 288 m/s

I don't know how to set up the equations to find the speed of the transverse waves on the string when the number of loops is equal to 8.

I keep getting the same velocity..

λ₈= 0.3m
f₈= 960hz

V= f₈ x λ₈ = 288 m/s ?

 

[anathema]

Thank you for posting your question. To solve part 6(c) of the problem, we can use the equation V = √(F/(m/L)), where V is the speed of the transverse wave, F is the tension force on the string, m is the mass of the string, and L is the length of the string. We know that the speed of the transverse wave is 288m/s, the tension force is 8.2944N, and the length of the string is 0.6m. We can rearrange the equation to solve for the mass of the string, which is equal to F/(V^2L).

Substituting in the values, we get m = 8.2944N/(288m/s)^2(0.6m) = 0.0001kg. Now, we can use this mass to find the speed of the transverse wave when the number of loops is equal to 8. We know that the mass does not change, but the length of the string does. Since the length of the string is halved (from 0.6m to 0.3m), the speed of the transverse wave will also be halved.

Therefore, the speed of the transverse wave when the number of loops is equal to 8 is 144m/s. This is different from the initial speed of 288m/s, as the speed of the wave is affected by the length of the string. I hope this helps you solve the problem. Good luck!