A particle executing SHM has a maximum speed of $30 c m s^{- 1}$ and a maximum acceleration of $60 c m s^{- 2}$ . The period of oscillation is:

π

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

\frac{π}{2} s

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

2 π

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

\frac{2 π}{15} s

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

Open in App

Solution

The correct option is C $\frac{2 π}{15} s$
Maximum velocity $V_{m a x} = A w = 30 c m s^{- 1}$
Maximum acceleration $a_{m a x} = A^{2} w = 60 c m s^{- 1}$
Where $A = A m p l i t u d e w = a n g u l a r f r e q u e n c y$
$∴ A^{2} w = 60$
$⟹ A \times 30 = 60 ⟹ A = 2 c m$
$∴ w = \frac{30}{2} s^{- 1} = 15 s e c^{- 1}$
$∴ T = \frac{2 π}{w} = \frac{2 π}{15} s e c$

Suggest Corrections

Similar questions

0.5 m s^{- 1}

1 m s^{- 2}

2 π {m/s}^{2}

6 m/s

2 π {m/s}^{2}

6 m/s

S H M

1 m / s

1.57 m / s^{2}

1 m s^{- 1}

1.57 m s^{- 2}

Join BYJU'S Learning Program