A particle is subjected to two simple harmonic motions of the same frequency and direction. The amplitude of the first motion is $4.0$ $c m$ and that of the second is $3.0$ $c m$ . Find resultant amplitude when the phage difference is $180^{0}$ . Find the resultant amplitude if the phase difference between the two motions is $0^{\circ}$ .

5 c m

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4 c m

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7 c m

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9 c m

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Solution

The correct option is B $7 c m$
In sucb situation, amplitudes are added by vector method.
$A_{R} = \sqrt{{(4)}^{2} + {(3)}^{2} + 2 (4) (3) cos ϕ}$
$= \sqrt{25 + 24 cos 0^{\circ}}$
$= 7$ cm

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The correct option is B 7cmIn sucb situation, amplitudes are added by vector method.AR=√(4)2+(3)2+2(4)(3)cosϕ =√25+24cos0∘ =7 cm

The correct option is B $7 c m$
In sucb situation, amplitudes are added by vector method.
$A_{R} = \sqrt{{(4)}^{2} + {(3)}^{2} + 2 (4) (3) cos ϕ}$
$= \sqrt{25 + 24 cos 0^{\circ}}$
$= 7$ cm