JEE Physics 2026 Question with Solution

Given: $m_1=1\,\text{kg}$, $m_2=0.2\,\text{kg}$, and $k=150\,\text{N/m}$ on a frictionless horizontal surface.

Find: The angular frequency of oscillation of the two-mass spring system.

For two masses connected by a spring on a frictionless surface, use the reduced mass:

$$\mu=\frac{m_1m_2}{m_1+m_2}$$

Then the angular frequency of relative oscillation is

$$\omega=\sqrt{\frac{k}{\mu}}$$

Substitute the given values:

$$\mu=\frac{(1)(0.2)}{1+0.2}=\frac{0.2}{1.2}=\frac{1}{6}\,\text{kg}$$

Now,

$$\omega=\sqrt{\frac{150}{1/6}}=\sqrt{900}=30$$

Equivalently,

$$\omega=\sqrt{\frac{k(m_1+m_2)}{m_1m_2}}=\sqrt{\frac{150\times1.2}{0.2}}=\sqrt{900}=30$$

Therefore, the angular frequency of the system is $30$ in SI units, so the correct option is D.

The solution also shows an intermediate statement claiming option B, but the worked calculation concludes $30$, so the final derived answer is taken as D.

Given: $m_1=1\,\text{kg}$, $m_2=0.2\,\text{kg}$, $k=150\,\text{N/m}$.

Find: The system angular frequency.

A direct standard result for two masses connected by a spring is

$$\omega=\sqrt{\frac{k(m_1+m_2)}{m_1m_2}}$$

Using the values,

$$\omega=\sqrt{\frac{150(1+0.2)}{(1)(0.2)}}$$ $$\omega=\sqrt{\frac{150\times1.2}{0.2}}=\sqrt{900}=30$$

Therefore, the correct option is D.