$\omega = \sqrt{\frac{k}{m}}$
$$\omega' = \sqrt{\frac{k/2}{2m}} = \sqrt{\frac{k}{4m}} = \frac{\omega}{2}$$
Back to JEE Main 2026 Waves & Oscillations
JEE Main 2026 — Physics Waves & Oscillations
hard
mcq
2026
Official previous-year question
Verified 30 May 2026.
Question
A spring-mass system oscillates with angular frequency $\omega$. If the mass is doubled and the spring constant is halved, the new angular frequency is:
Options
- A
$\frac{\omega}{2}$
- B
$\frac{\omega}{\sqrt{2}}$
- C
$2\omega$
- D
$\omega\sqrt{2}$
Solution
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