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:

$$\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}$$