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. $3,4$

Answer 1

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$10$