The displacement of a particle is given by $y = 0.25 sin 200 t c m$ . The maximum speed of the particle is :

200 c m s^{- 1}

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

100 c m s^{- 1}

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

50 c m s^{- 1}

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

5.25 c m s^{- 1}

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

Open in App

Solution

The correct option is C $50 c m s^{- 1}$
The displacement of he particle is given by
$y = 0.25 s i n 200 t$
velocity, $v = \frac{d y}{d t}$
$v = 0.25 \times 200 c o s 200 t$
$v = 50 c o s 200 t$
The maximum speed of the particle is $50 c m / s .$
The correct option is C.

Suggest Corrections

Similar questions

Q. The displacement of a particle executing SHM is given y = 0.25 sin 200t cm. The maximum speed of the particle is

Q. The displacement of a particle of a string carrying a travelling wave is given by

y = 3 sin [6.28 (0.50 x - 50 t)]

(x \to in cm, y \to cm, t \to in second)

.
What would be the speed of the wave?

Q. The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s⁻¹ and 50 cm s⁻². Find the position(s) of the particle when the speed is 8 cm s⁻¹.

The maximum speed and acceleration of a particle executing simple harmonic motion are $10 c m s^{- 1} a n d 50 c m s^{- 2}$ . Find the position(s) of the particle when the speed is $8 c m^{- 1}$ .

Q. The motion a particle is given by

y = U t + \frac{1}{2} {g t}^{2}

. Whose

y

is the displacement along

y

is calculate what kind of acceleration the particle is going through.

Join BYJU'S Learning Program

The displacement of a particle is given by y=0.25sin200 t cm. The maximum speed of the particle is :

The correct option is C 50 cm s−1The displacement of he particle is given byy=0.25sin200tvelocity, v=dydtv=0.25×200cos200tv=50cos200tThe maximum speed of the particle is 50cm/s.The correct option is C.

The displacement of a particle is given by $y = 0.25 sin 200 t c m$ . The maximum speed of the particle is :

The correct option is C $50 c m s^{- 1}$
The displacement of he particle is given by
$y = 0.25 s i n 200 t$
velocity, $v = \frac{d y}{d t}$
$v = 0.25 \times 200 c o s 200 t$
$v = 50 c o s 200 t$
The maximum speed of the particle is $50 c m / s .$
The correct option is C.