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Questions  

The phase difference between two SHM y1 = 10 sin (10πt + π3) and y2 = 12 sin (8πt+π4) at t = 0.5 s is

 

a
11π12
b
13π12
c
π
d
17π12

detailed solution

Correct option is B

Phase for 1st SHM: ∅ = 10πt+π3Phase for 2nd SHM: ∅ 2 = 8πt+π4Phase difference: ∆∅ = ∅1-∅2 = 2πt + π12Put t = 0.5 s∆∅ = 2π(0.5)+π12 = 13π12

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At time t = 0, one particle is at maximum positive amplitude and the other particle is at half of the positive amplitude. Their amplitudes and time periods T are same. If they are approaching, find the time by which they cross each other:


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