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Physics Question 35 – JEE-MAIN 2026

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Match List - I with List - II. List - IList - IIA. sin2⁡(ωt)I. Periodic with time period T=πω but not simple harmonic motion (SHM)B. sin3⁡(2ωt)II. Periodic with time period T=2πω but Not SHMC. sin⁡(ωt)+cos⁡(πωt)III. Periodic with time period T=πω and SHMD. cos⁡(ωt)+cos⁡(2ωt)IV. Non-periodic Choose the correct answer from the options given below:

Simplify the given functions using relevant trigonometric identities to identify their fundamental frequencies and periods.

● Video Walkthrough
Step 1: Analyze Function A: sin2⁡(ωt)✦ Active

Using the identity sin2⁡(x)=1−cos⁡(2x)2, we can rewrite the function as:

sin2⁡(ωt)=1−cos⁡(2ωt)2=12−12cos⁡(2ωt)

This represents a simple harmonic motion (SHM) about a mean position of 12 with angular frequency 2ω. The time period is T=2π2ω=πω. Thus, A matches with III.

Step 2: Analyze Functions B, C, D○ Expand

For B. sin3⁡(2ωt): Using sin3⁡(x)=3sin⁡(x)−sin⁡(3x)4, we get 34sin⁡(2ωt)−14sin⁡(6ωt). This is a superposition of two SHMs with periods T1=2π2ω=πω and T2=2π6ω=π3ω. The fundamental period is LCM(πω,π3ω)=πω. It is not SHM. Thus, B matches with I.

For C. sin⁡(ωt)+cos⁡(πωt): The periods are T1=2πω and T2=2ππω=2ω. The ratio T1T2=2π/ω2/ω=π, which is irrational. Therefore, the function is non-periodic. Thus, C matches with IV.

For D. cos⁡(ωt)+cos⁡(2ωt): This is a superposition of two SHMs with periods T1=2πω and T2=2π2ω=πω. The fundamental period is LCM(2πω,πω)=2πω. It is not SHM. Thus, D matches with II.

💡 Teacher's Secret Hint

Remember that a superposition of SHMs with different frequencies is periodic but not SHM.

Step 3: Final Matching and Option Selection○ Expand

Combining the matches:

A - III

B - I

C - IV

D - II

This combination corresponds to option 1.

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