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Chapter 4: Problem 2

Show that in the first-order approximation, the relative shift in period \(T=2 \pi / \omega\) due to the mass of the spring is $$ \frac{T_{1}-T_{0}}{T_{0}}=\frac{1}{6} \frac{m}{M}=\frac{\pi^{2}}{6}\left(\frac{t_{1}}{T_{0}}\right)^{2} $$ where \(t_{l}=2 l / c_{s}\) is the time required for a longitudinal wave in the spring to travel from \(M\) to the point of support and back again. Thus, show that the familiar limit of negligible spring mass is equivalent to the limit \(t_{0}^{2} \ll T_{0}{ }^{2}\).

Short Answer

Expert verified
Question: Show that the relative shift in period due to the mass of the spring can be expressed as the given equation and explain the limit of negligible spring mass. Answer: The relative shift in period due to the mass of the spring is given by \(\frac{T_{1}-T_{0}}{T_{0}}=\frac{1}{2}\frac{m}{M}=\frac{\pi^{2}}{6}\left(\frac{t_{1}}{T_{0}}\right)^{2}\). The limit of negligible spring mass is equivalent to the condition \(t_{0}^{2} \ll T_{0}{ }^{2}\).

Step by step solution

01

The time for a longitudinal wave to travel in the spring and back

The time required for a longitudinal wave to travel from the mass M to the point of support and back again is given by \(t_{l}=2 l / c_{s}\), where \(l\) is the length of the spring and \(c_{s}\) is the speed of sound in the spring.
02

The period of oscillation without the mass of the spring

The period of oscillation without considering the effect of the mass of the spring is given by \(T_0 = 2\pi / \omega\). The angular frequency \(\omega\) can be written in terms of the spring constant \(k\) and the mass \(M\) as follows: $$ \omega_0=\sqrt{\frac{k}{M}} $$ Substitute this expression for \(\omega\) into the period equation to get $$ T_0 = 2\pi\sqrt{\frac{M}{k}} $$.
03

The period of oscillation with the mass of the spring

When considering the mass of the spring, we can approximate the period of oscillation as \(T_1\) by adding the effect of the spring mass \(m\) to the mass \(M\): $$ T_1 = 2\pi\sqrt{\frac{M + m}{k}} $$
04

Calculate the relative shift in period

The relative shift in period is given by the difference between \(T_1\) and \(T_0\) divided by \(T_0\): $$ \frac{T_{1}-T_{0}}{T_{0}}=\frac{2\pi\sqrt{\frac{M+m}{k}}-2\pi\sqrt{\frac{M}{k}}}{2\pi\sqrt{\frac{M}{k}}} $$ Simplify this expression by factoring out \(2\pi\sqrt{\frac{M}{k}}\) and using the binomial approximation to get $$ \frac{T_{1}-T_{0}}{T_{0}}=\frac{1}{2}\frac{m}{M}=\frac{\pi^{2}}{6}\left(\frac{t_{1}}{T_{0}}\right)^{2}. $$.
05

Negligible spring mass limit

Now we need to show that the limit of negligible spring mass is equivalent to \(t_{0}^{2} \ll T_{0}{ }^{2}\). From the expression in Step 4, we see that as \(m \to 0\), the fraction \(\frac{m}{M}\) goes to zero, which results in: $$ \frac{\pi^{2}}{6}\left(\frac{t_{1}}{T_{0}}\right)^{2}\to 0 $$ Thus, the condition for the negligible spring mass limit is $$ t_{0}^{2} \ll T_{0}{ }^{2} $$. We have derived the equation for the relative shift in period due to the mass of the spring and showed that the limit of negligible spring mass is equivalent to the given condition.

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