Oscillations

JEE Main level, single correct, medium

127 of 194

Question 127 of 194

An object of mass mm is suspended at the end of a massless wire of length LL and area of cross-section, A. Young modulus of the material of the wire is YY. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is :

Show answer and solution

Answer: (B) f=12πYAmLf=\dfrac{1}{2\pi }\sqrt{\dfrac{YA}{mL}}

f=12πkm=12π(YAmL)f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}=\dfrac{1}{2\pi }\left(\sqrt{\dfrac{YA}{mL}}\right)

f=12πYAmLf=\dfrac{1}{2\pi }\sqrt{\dfrac{YA}{mL}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Simple pendulum simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Oscillations (Time Period and Frequency), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.