Question

The equation of motion for an oscillating particle is given by x = 3 sin 4$\pi$t + 4 cos 4$\pi$t, where x is in mm and t is in second. The particle

• A. starts its motion from rest.
• B. starts its motion with an initial velocity u = 12$\pi$ mm/s
• C. has its maximum acceleration equal to 80${\pi }^2$ mm/$s^2$
• D. has its maximum velocity equal to 20$\pi$ mm/s

Answer 1

Answer: (B), (C), (D)

The correct options are
B starts its motion with an initial velocity u = 12$\pi$ mm/s
C has its maximum acceleration equal to 80${\pi }^2$ mm/$s^2$
D has its maximum velocity equal to 20$\pi$ mm/s

Given: $x=3\sin 4\pi t+4\cos 4\pi t$

For simplifying, Multipying and dividing equation by $\sqrt{3^2+4^2}=5$, we get

$x=5(\frac{3}{5}\sin 4\pi t+\frac{4}{5}\cos 4\pi t)$

Let, $\cos \varphi =\frac{3}{5}$, then $\sin \varphi =\frac{4}{5}$

$x=5(\cos \varphi \sin 4\pi t+\sin \varphi \cos 4\pi t)=5\sin (4\pi t+\varphi )$

Now, velocity of particle at any time t is given by, $v=\frac{dx}{dt}$

$v=\frac{d}{dt}\left(5\sin (4\pi t+\varphi )\right)=20\pi \cos (4\pi t+\varphi )$

Acceleration of particle at any time t is given by, $a=\frac{dv}{dt}$

$a=\frac{d}{dt}\left(20\pi \cos (4\pi t+\varphi )\right)=80{\pi }^2(-\sin (4\pi t+\varphi ))$

For initial velocity, $t=0$

$v_0=12\pi \cos 0-16\pi \sin 0=12\pi$ $mm/s$

Velocity will be maximum when $\cos (4\pi t+\varphi )$ will be maximum, i.e, $1$

So, $v_{max}=20\pi$ $mm/s$

Acceleration will be maximum when $\sin (4\pi t+\varphi )$ will be maximum, i.e, $1$,

so $a_{max}=80{\pi }^2$ $mm/s^2$