Question 175: Oscillations

Question

A mass $$M$$ is 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 $${{5T} \over 3}$$. Then the ratio of $${{m} \over M}$$ is

Accepted Answer

The time period of a simple harmonic motion (SHM) performed by a mass-spring system is given by the formula:
$$T = 2\pi \sqrt{{M \over k}}$$
where:
T is the time period,
M is the mass of the object, and
k is the spring constant.
We know that if the mass is increased by m, the time period becomes $\frac{5T}{3}$. We can set up an equation for this new scenario:
$$\frac{5T}{3} = 2\pi \sqrt{\frac{M + m}{k}}$$
Since we know that $T = 2\pi \sqrt{\frac{M}{k}}$, we can substitute T in the equation above:
$$\frac{5}{3} \cdot 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M + m}{k}}$$
Squaring both sides of the equation to eliminate the square root, we get:
$$\frac{25}{9} \cdot \frac{M}{k} = \frac{M + m}{k}$$
Solving for $\frac{m}{M}$, we get:
$$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$$

Answer

$${3 \over 5}$$

Answer

$${25 \over 9}$$

Answer

$${16 \over 9}$$

Answer

$${5 \over 3}$$

Oscillations

A mass $M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m. the time period becomes {{5T} \over 3}. Then the ratio of {{m} \over M}$ is

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