You visited us 1 times! Enjoying our articles? Unlock Full Access!

Second Law of Motion

The displacem...

Question

The displacement of a particle executing S.H.M is given by $x$ = $0.25 s i n 200 t$ in meters, the maximum speed of particle is:

50 m / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

25 m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200 m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $50 m / s$
$x = 0.25 s i n 200 t$
Speed, $v = \frac{d x}{d t} = 0.25 \times 200 c o s 200 t$
The maximum speed of particle $v_{m} = 50 m / s$
The correct option is A.

Suggest Corrections

Similar questions

S . H . M .

10 c m / s

8 c m / s

4 c m

5 c m

v = 6 t^{2} + 2 t + 5 m/s

4 sec

4 sec

x

v

4 v^{2} = 25 - x^{2}

s = 18 t + 3 t^{2} - 2 t^{3}

s

t

π

Join BYJU'S Learning Program