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Chapter 9: Problem 10

A mass \(M\) suspended from a spring of negligible mass. The spring is pulled a little and then released, so that the mass executes SHM of time period \(T\). If the mass is increased by \(m\), the time period becomes \(5 T / 3\). The ratio of \(m / M\) is (A) \(\frac{5}{3}\) (B) \(\frac{3}{5}\) (C) \(\frac{16}{9}\) (D) \(\frac{25}{9}\)

Short Answer

Expert verified
The ratio of \(m/M\) is \(\boxed{\frac{16}{9}}\) (Option C).

Step by step solution

01

Write down the formula for the time period of SHM in a mass-spring system

The time period of SHM for a mass suspended from a spring is given by: \[ T = 2\pi\sqrt{\frac{M}{k}} \] where \(T\) is the time period, \(M\) is the mass, and \(k\) is the spring constant.
02

Write the formula for the time period with increased mass

When mass \(m\) is added to the system, the total mass becomes \(M+m\). The time period in this case becomes \(5T/3\). Therefore, \[ \frac{5T}{3} = 2\pi\sqrt{\frac{M+m}{k}} \]
03

Set up a system of equations to solve for the ratio m/M

Now we have two equations with two unknowns, \(m\) and \(k\): \[ \begin{cases} T = 2\pi\sqrt{\frac{M}{k}} & \\ \frac{5T}{3} = 2\pi\sqrt{\frac{M+m}{k}} & \end{cases} \]
04

Solve the system of equations

Let's divide the second equation by the first equation to eliminate the unknown \(k\): \[ \frac{\frac{5T}{3}}{T} = \frac{2\pi\sqrt{\frac{M+m}{k}}}{2\pi\sqrt{\frac{M}{k}}} \] This simplifies to: \[ \frac{5}{3} = \sqrt{\frac{M+m}{M}} \]
05

Solve for the ratio \(m/M\)

Square both sides of the equation to get rid of the square root: \[ \left(\frac{5}{3}\right)^2 = \frac{M+m}{M} \] Now we can solve for the ratio \(m/M\): \[ \frac{m}{M} = \left(\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{16}{9} \] Therefore, the ratio of \(m/M\) is \(\boxed{\frac{16}{9}}\) (Option C).

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