First slide

Simple hormonic motion

Question

The phase difference between two SHM $y_{1} = 10 \sin (10 πt + \frac{π}{3}) and y_{2} = 12 \sin (8 πt + \frac{π}{4}) at t = 0.5 s is$

Easy

A

$\frac{11 π}{12}$
B

$\frac{13 π}{12}$
C

$π$
D

$\frac{17 π}{12}$

Solution

Phase for 1st SHM: $\emptyset = 10 πt + \frac{π}{3}$
Phase for 2nd SHM: $\emptyset_{2} = 8 πt + \frac{π}{4}$
Phase difference: $∆ \emptyset = \emptyset_{1} - \emptyset_{2} = 2 πt + \frac{π}{12}$
Put t = 0.5 s
$∆ \emptyset = 2 π (0.5) + \frac{π}{12} = \frac{13 π}{12}$

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