The piston in the cylinder head of a locomotive has a stroke of $6 m$ . If the piston executing simple harmonic motion with an angular frequency of $200 r a d m i n^{- 1}$ , its maximum speed is

5 m s^{- 1}

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

10 m s^{- 1}

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

15 m s^{- 1}

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

20 m s^{- 1}

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

Open in App

Solution

The correct option is B $10 m s^{- 1}$
Here, $ω = 200 r a d m i n^{- 1}$
Stroke means twice the amplitude.
$∴ A = 3 m$ ,
Maximum speed,
$v_{m a x} = A ω = 3 \times 200 = 600 m r a d m i n^{- 1} = \frac{600}{60} m s^{- 1} = 10 m s^{- 1}$ .

Suggest Corrections

Similar questions

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

Q. A particle executing simple harmonic motion has an angular frequency of

6.28 s^{- 1}

and amplitude of

10 c m

. The maximum speed of the particle is :

Q. A particle executing simple harmonic motion has an angular frequency of

6.28 s^{- 1}

and amplitude of

10 c m

. The maximum acceleration of the particle is :

A particle executing simple harmonic motion has an angular frequency of 6.28

s^{- 1}

and amplitude of 10 cm, find the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.

Q. A particle executing simple harmonic motion has an angular frequency of

6.28 s^{- 1}

and amplitude of

10 c m

. The time period of the SHM is :

Join BYJU'S Learning Program

The piston in the cylinder head of a locomotive has a stroke of 6 m. If the piston executing simple harmonic motion with an angular frequency of 200 rad min−1, its maximum speed is

The correct option is B 10 ms−1Here, ω=200radmin−1Stroke means twice the amplitude.∴A=3m,Maximum speed, vmax=Aω=3×200=600mradmin−1=60060ms−1=10ms−1.

The piston in the cylinder head of a locomotive has a stroke of $6 m$ . If the piston executing simple harmonic motion with an angular frequency of $200 r a d m i n^{- 1}$ , its maximum speed is

The correct option is B $10 m s^{- 1}$
Here, $ω = 200 r a d m i n^{- 1}$
Stroke means twice the amplitude.
$∴ A = 3 m$ ,
Maximum speed,
$v_{m a x} = A ω = 3 \times 200 = 600 m r a d m i n^{- 1} = \frac{600}{60} m s^{- 1} = 10 m s^{- 1}$ .