Two particles $A$ and $B$ execute simple harmonic motion of period $T$ and $\frac{5T}{4}$. They start from mean position. The phase difference between them when the particle $A$ complete an oscillation will be

Question

Dr. Abubacker · Accepted Answer

(a) Given that, the time period of particle $A = T$ and the time period of particle $B = \frac{5T}{4}$
Hence, the time difference $(\mathrm{\Delta}T) = \frac{5T}{4} - T$
$\Rightarrow \ \ \ \ \ \ \ \ \mathrm{\Delta}T = \frac{T}{4}$ …(i)
The relation between phase difference and time difference is
$\mathrm{\Delta}\varnothing = \frac{2\pi}{T}\mathrm{\Delta}T$
$\mathrm{\Delta}\varnothing = \frac{2\pi}{T} 	imes \frac{T}{4}$
$\Rightarrow \ \ \ \ \ \ \ \ \mathrm{\Delta}\varnothing = \frac{\pi}{2}$

Q. Two particles $A$ and $B$ execute simple harmonic motion of period $T$ and $\frac{5T}{4}$. They start from mean position. The phase difference between them when the particle $A$ complete an oscillation will be

Explanation

Reviewed by Dr. Abubacker

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