Q.35. Match List-I with List-II.List-I: A. , B. , C. , D. . List-II: I. Periodic with time period b

Q: What is the correct answer for JEE Main Physics Q35?

The correct answer is A. A. $\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$. This represents SHM with period $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. So A-III. B. $\sin^3(2\omega t) = \frac{3}{4} \sin(2\omega t) - \frac{1}{4} \sin(6\omega t)$. This is a superposition of two SHMs with periods $\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ and $\frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$. The common period is $\frac{\

Q: Q.35. Match List-I with List-II.List-I: A. , B. , C. , D. . List-II: I. Periodic with time period but not simple harmonic motion (SHM), II. Periodic with time period but Not SHM, III. periodic with…

The correct answer is A. A. $\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$. This represents SHM with period $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. So A-III. B. $\sin^3(2\omega t) = \frac{3}{4} \sin(2\omega t) - \frac{1}{4} \sin(6\omega t)$. This is a superposition of two SHMs with periods $\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ and $\frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$. The common period is $\frac{\

Q: What are the options for this JEE Main Physics question?

A. A-III, B-I, C-IV, D-II · B. A-II, B-I, C-III, D-IV · C. A-III, B-II, C-IV, D-I · D. A-II, B-I, C-IV, D-III

Question

Q.35. Match List-I with List-II.List-I: A. , B. , C. , D. . List-II: I. Periodic with time period  but not simple harmonic motion (SHM), II. Periodic with time period  but Not SHM, III. periodic with time period  and SHM, IV. Non-periodic. Choose the correct answer from the options given below.

FirstInTest · Accepted Answer

A. $\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$. This represents SHM with period $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. So A-III.
B. $\sin^3(2\omega t) = \frac{3}{4} \sin(2\omega t) - \frac{1}{4} \sin(6\omega t)$. This is a superposition of two SHMs with periods $\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ and $\frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$. The common period is $\frac{\pi}{\omega}$, but it's not simple harmonic motion. So B-I.
C. $\sin(\omega t) + \cos(\pi \omega t)$. The frequencies are $\omega$ and $\pi \omega$. Since $\pi$ is irrational, there is no common period, so it's non-periodic. So C-IV.
D. $\cos \omega t + \cos 2\omega t$. The periods are $\frac{2\pi}{\omega}$ and $\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. The common period is $\frac{2\pi}{\omega}$, but it does not represent SHM. So D-II.

Answer

A-III, B-I, C-IV, D-II

Answer

A-II, B-I, C-III, D-IV

Answer

A-III, B-II, C-IV, D-I

Answer

A-II, B-I, C-IV, D-III

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