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Chapter 14: Problem 59

An ideal spring is stretched to a total length \(L_{1} .\) When that length is doubled, the speed of transverse waves on the spring triples. Find an expression for the unstretched length of the spring.

Short Answer

Expert verified
The unstretched length of the spring is \( \frac{1}{9} L_{1} \).

Step by step solution

01

Write Out the Two Equations From the Problem Statement

The speed of a wave on a stretched string or spring is given by \(v = \sqrt{T / \mu}\). For the first case, we have \(v_{1} = \sqrt{T_{1} / \mu_{1}}\). For the second case, where the length is doubled and the speed is tripled, we have \(v_{2} = 3 * v_{1} = \sqrt{T_{2} / \mu_{2}}\). We also know that \(\mu = m / L\). So, we can rewrite these equations as \(v_{1} = \sqrt{T_{1} / (m / L_{1})}\) and \(3 * v_{1} = \sqrt{T_{2} / (m / (2 * L_{1}))}\).
02

Rewrite the Second Equation

We can rewrite the second equation in terms of \(v_{1}\) and \(L_{1}\). We do this by squaring both sides of the equation and isolating \(L_{1}\). We get \(L_{1} = \frac{4 * T_{2}}{(3 * v_{1})^2 * m}\).
03

Solve for the Unstretched Length

The unstretched length of the spring, \(L_{0}\), can be found by considering the change in the spring's length when a tension is applied. If we call \(T_{1}\) the tension when the spring's length is \(L_{1}\), and \(T_{2}\) the increased tension when the length is doubled, we can write \(L_{1} = L_{0} + x_{1}\) and \(2 * L_{1} = L_{0} + x_{2}\). From this we get \(L_{0} = 2 * L_{1} - x_{2}\). Since \(x_{2} = 2 * L_{1}\), we sub in and simplify to get \(L_{0} = 2 * L_{1} - 2 * L_{1} = 0\).
04

Solve for Tension under Doubled Length

We also know from Hook's law that spring tension is proportional to stretch distance 鈥� \(T_{2} = k * x_{2}\) and \(T_{1} = k * x_{1}\). Hence \(T_{2} = 2 * T_{1}\). Substituting this into the equation we found in Step 2 gives \(L_{1} = \frac{4 * (2 * T_{1})}{9 * T_{1} * m / L_{1}} = \frac{8}{9} L_{1}\), which simplifies to \(L_{0} = \frac{1}{9} L_{1}\). Hence, the unstretched length of the spring is one ninth of its initial stretched length.

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Most popular questions from this chapter

When a \(340-\mathrm{g}\) spring is stretched to a total length of \(40 \mathrm{cm},\) it supports transverse waves propagating at \(4.5 \mathrm{m} / \mathrm{s} .\) When it's stretched to \(60 \mathrm{cm},\) the waves propagate at \(12 \mathrm{m} / \mathrm{s}\). Find (a) the spring's unstretched length and (b) its spring constant.

Two waves have the same angular frequency \(\omega,\) wave number \(k\) and amplitude \(A,\) but they differ in phase: \(y_{1}=A \cos (k x-\omega t)\) and \(y_{2}=A \cos (k x-\omega t+\phi) .\) Show that their superposition is also a simple harmonic wave, and determine its amplitude as a function of the phase difference \(\phi.\)

A rope is stretched between supports \(18.3 \mathrm{m}\) apart; its tension is \(78.6 \mathrm{N} .\) If one end of the rope is tweaked, the resulting disturbance reaches the other end 585 ms later. Find the rope's mass.

If you place a perfectly clear piece of glass in perfectly clear water, you can still see the glass. Why?

If you doubled the tension in a string, what would happen to the speed of waves on the string?
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