Question

# The displacement equation of a particle is $$x = 3 \sin 2t + 4 \cos 2t$$ . The amplitude and maximum velocity will be respectively

A
5,10
B
3,2
C
4,2
D
3,4

Solution

## The correct option is A $$5,10$$The displacement of the particle is in the form of $$x = A\sin \omega t + B\cos \omega t$$ The amplitude is given as $$A = \sqrt {{3^2} + {4^2}}$$ Hence the amplitude is given as$$5$$ The phase difference is of$$\frac{\pi }{2}$$ The maximum speed is given as $${v_{\max }} = A\omega$$ $${v_{\max }} = 5 \times 2$$ Hence the maximum velocity is$$10$$Physics

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