Oscillations

JEE Main level, single correct, medium

135 of 194

Question 135 of 194

A particle executes simple harmonic motion and it is located at x=a,bx=a,b and cc at time t0,2t0and3t0{t}_{0},2{t}_{0}and3{t}_{0} respectively. The frequency of the oscillation is:

Show answer and solution

Answer: (A) 12πt0cos⁡−1⁡(a+c2b)\dfrac{1}{2\pi {t}_{0}}{\cos }^{-1}⁡\left(\dfrac{a+c}{2b}\right)

a=Asin⁡⁡ωt0a=A\sin ⁡\omega {t}_{0}...(1)(1)
b=Asin⁡⁡2ωt0b=A\sin ⁡2\omega {t}_{0}...(2)(2)
c=Asin⁡⁡3ωt0c=A\sin ⁡3\omega {t}_{0}...(3)(3)
adding equation (1)(1) and (3)(3)

a+c=A[sin⁡⁡ωt0+sin⁡⁡3ωt0]=2Asin⁡⁡2ωt0cos⁡⁡ωt0a+c=A\left[\sin ⁡\omega {t}_{0}+\sin ⁡3\omega {t}_{0}\right]=2A\sin ⁡2\omega {t}_{0}\cos ⁡\omega {t}_{0}...(4)(4)

∵sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2∵ \sin C+\sin D= 2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}
from equation (2)(2) and (4)(4)
a+cb=2cos⁡ωt0\dfrac{a+c}{b}=2\cos \omega {t}_{0}

ω=1t0cos⁡−1⁡(a+c2b)⇒f=12πt0cos⁡−1⁡(a+c2b)\omega =\dfrac{1}{{t}_{0}}{\cos }^{-1}⁡\left(\dfrac{a+c}{2b}\right)\Rightarrow f=\dfrac{1}{2\pi {t}_{0}}{\cos }^{-1}⁡\left(\dfrac{a+c}{2b}\right)

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.