Home Courses Contact About


Topic 09 : Super Position of SHM

1. ⇒ (NEET 2019)

The displacement of a particle executing simple harmonic motion is given by

y = A0 + A sin ω t + B cos ω t.

Then the amplitude of its oscillation is given by :

A. A 2 + B 2

B. A+B

C. A + A 2 + B 2

D. A 0 2 + ( A + B ) 2

Correct Answer is Option (A)

From the given displacement

y = A0 + A sin ω t + B cos ω t.

Let assume, y - A0 = γ

γ = A sin ω t + B cos ω t

= A 2 + B 2 sin ⁡ ( ω t + ϕ )

which is S.H.M

∴ Resultant amplitude of the particle,

= A 2 + B 2

2. ⇒ (AIPMT 2015 Cancelled Paper)

When two displacements represented by y1 = a sin ( ω t ) and y2 = b cos ( ω t ) aresuperimposed the motion is :

A. simple harmonic with amplitude a 2 + b 2

B. simple harmonic with amplitude ( a + b ) 2

C. not a simple harmonic

D. simple harmonic with amplitude a b

Correct Answer is Option (A)

Here, y 1 = a sin ⁡ ω t

y 2 = b cos ⁡ ω = b sin ⁡ ( ω t + π 2 )

AIPMT 2015 Cancelled Paper Physics - Oscillations Question 41 English Explanation
Hence, resultant motion is SHM with amplitude a 2 + b 2 .