A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) $0^{\circ},$ (b) $60^{\circ},$ (c) $90^{\circ} .$

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Solution

The particle is subjected to two SHMs of same time period, in the same direction.

Given, $r_{1} = 3 c m, r_{2} = 4 c m$

and $Φ$ = phase difference Resultant amplitude

$= R = \sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s Φ}$

(a) When $Φ = 0^{\circ}$

$R = \sqrt{(3^{2} + 4^{2} + 2.3.4 c o s 0^{\circ})}$

$= 7 c m$

(b) When $Φ = 60^{\circ}$

$R = \sqrt{(3^{2} + 4^{2} + 2.3.4 c o s 60^{\circ})}$

$= \sqrt{37} = 6.1 c m$

(c) When $Φ = 90^{\circ}$

$R = \sqrt{(3^{2} + 4^{2} + 2.3.4 c o s 90^{\circ})}$

$= \sqrt{25} = 5 c m$

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A particle is subjected to two simple harmonic motions of same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0∘, (b) 60∘, (c) 90∘.