If transverse waves on a string travel with a velocity of \(50 \mathrm{~m} / \mathrm{s}\) when the string is under a tension of \(20 \mathrm{~N},\) what tension on the string is required for the waves to travel with a velocity of \(30 \mathrm{~m} / \mathrm{s} ?\) a) \(7.2 \mathrm{~N}\) c) \(33 \mathrm{~N}\) e) \(45 \mathrm{~N}\) b) \(12 \mathrm{~N}\) d) \(40 \mathrm{~N}\) f) \(56 \mathrm{~N}\)

Short Answer

Expert verified
Answer: (a) 7.2 N

Step by step solution

01

Write down given information

We are given the following information: Initial wave speed \(v_1 = 50 \mathrm{~m} / \mathrm{s}\) Initial tension \(T_1 = 20 \mathrm{~N}\) Desired wave speed \(v_2 = 30 \mathrm{~m} / \mathrm{s}\) We need to find the desired tension \(T_2\).
02

Write down wave speed formula

We'll use the wave speed formula: \(v = \sqrt{\frac{T}{\mu}}\), where \(v\) is the wave speed, \(T\) is the tension, and \(\mu\) is the mass per unit length of the string.
03

Set up the relationship between initial and desired wave speeds and tension

We have the two wave speeds related to two different tensions, \(v_1 = \sqrt{\frac{T_1}{\mu}}\) and \(v_2 = \sqrt{\frac{T_2}{\mu}}\). Divide the first equation by the second equation to eliminate \(\mu\): \(\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}\)
04

Solve for desired tension

Now we have a relationship we can use to find \(T_2\). First, square both sides of the equation: \(\frac{v_1^2}{v_2^2}=\frac{T_1}{T_2}\) Next, solve for \(T_2\): \(T_2 = T_1 \cdot \frac{v_2^2}{v_1^2}\) Plug the given values for \(v_1\), \(v_2\), and \(T_1\): \(T_2 = 20 \mathrm{N} \cdot \frac{(30 \mathrm{~m} / \mathrm{s})^2}{(50 \mathrm{~m} / \mathrm{s})^2}\) Solve for \(T_2\): \(T_2 = 20 \mathrm{N} \cdot \frac{900}{2500} \Rightarrow T_2 = 20 \mathrm{N} \cdot 0.36 \Rightarrow T_2 = 7.2 \mathrm{~N}\) The required tension for the waves to travel with a velocity of \(30 \mathrm{~m} / \mathrm{s}\) is \(7.2 \mathrm{~N}\), which corresponds to answer (a).

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Most popular questions from this chapter

Consider a guitar string stretching \(80.0 \mathrm{~cm}\) between its anchored ends. The string is tuned to play middle \(\mathrm{C},\) with a frequency of \(256 \mathrm{~Hz}\), when oscillating in its fundamental mode, that is, with one antinode between the ends. If the string is displaced \(2.00 \mathrm{~mm}\) at its midpoint and released to produce this note, what are the wave speed, \(v\), and the maximum speed, \(V_{\text {max }}\), of the midpoint of the string?

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