Physics Q37 – JEE-MAIN 2026

The frequency of oscillation of a mass

m

suspended by a spring is

v_{1}

. If the length of the spring is cut to half, the same mass oscillates with frequency

v_{2}

. The value of

\frac{v_{2}}{v_{1}}

is _______.

Clue 1 of 3: Understand the relationship between the frequency of oscillation, mass, and spring constant in a simple harmonic motion (SHM) system

🧠 Full Solution Path

Step 1: Relate frequency to spring constant✦ Active

The frequency of oscillation ( $v$ ) of a mass $m$ suspended by a spring with spring constant $k$ is given by the formula:

v = \frac{1}{2 π} \sqrt{\frac{k}{m}}

For the initial state, $v_{1} = \frac{1}{2 π} \sqrt{\frac{k_{1}}{m}}$ .

Step 2: Determine change in spring constant○ Expand

The spring constant $k$ of a spring is inversely proportional to its length $L$ . This means $k L = constant$ . If the original length is $L_{1}$ and the spring constant is $k_{1}$ , then when the length is cut to half, the new length $L_{2} = \frac{L_{1}}{2}$ . The new spring constant $k_{2}$ will be:

k_{1} L_{1} = k_{2} L_{2} ⟹ k_{1} L_{1} = k_{2} (\frac{L_{1}}{2}) ⟹ k_{2} = 2 k_{1}

So, the spring constant doubles when the length is halved.

💡 Teacher's Secret Hint

Remember that cutting a spring effectively makes it 'stiffer'.

Step 3: Calculate the ratio of frequencies○ Expand

The new frequency $v_{2}$ with the spring constant $k_{2} = 2 k_{1}$ and the same mass $m$ is:

v_{2} = \frac{1}{2 π} \sqrt{\frac{k_{2}}{m}} = \frac{1}{2 π} \sqrt{\frac{2 k_{1}}{m}}

Now, we find the ratio $\frac{v_{2}}{v_{1}}$ :

\frac{v_{2}}{v_{1}} = \frac{\frac{1}{2 π} \sqrt{\frac{2 k_{1}}{m}}}{\frac{1}{2 π} \sqrt{\frac{k_{1}}{m}}} = \sqrt{\frac{2 k_{1} / m}{k_{1} / m}} = \sqrt{2}

The value of $\frac{v_{2}}{v_{1}}$ is $\sqrt{2}$ .

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Physics Question 37 – JEE-MAIN 2026

Clue 1 of 3: Understand the relationship between the frequency of oscillation, mass, and spring constant in a simple harmonic motion (SHM) system

🧠 Full Solution Path