For periodic motion of small amplitude $A$ , the time period $T$ of this particle is proportional to

A \sqrt{\frac{m}{α}}

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

\frac{1}{A} \sqrt{\frac{m}{α}}

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

A \sqrt{\frac{α}{m}}

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

A \sqrt{\frac{2 α}{m}}

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

Open in App

Solution

The correct option is A $\frac{1}{A} \sqrt{\frac{m}{α}}$
The dimensions of $α$ are $\frac{V (x)}{x^{4}} = \frac{E n e r g y}{{L e n g t h}^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$
It is clear that $\frac{1}{A} \sqrt{\frac{m}{α}}$ has dimensions of time.

Suggest Corrections

Similar questions

A balloon mass M is descending at a constant acceleration $α$ . When a mass m is released from the balloon. It starts rising with the same acceleration $α$ . What is the value of m?

Q. A body of mass M is situated in a potential field

U (x) = U_{0} (1 - c o s α x)

where

U_{0}

and

α

are constants. The time period of small oscillations is

Q. An

α

- particle travels in a circular path of radius 0.45 m in a magnitude field B = 1.2 Wb/

m^{2}

with a speed of 2.6 x

10^{7}

m/s. The period of revolution of the

α

- particle is:

A balloon of mass M is descending at a constant acceleration $α$ . When a mass m is released from the balloon, it starts rising with the same acceleration $α$ . Assuming that its volume does not change, what is the value of m?

Q. Assertion :If a proton and an

α

-particle enter a uniform magnetic field perpendicularly with the same speed, then the time period of revolution of the

α

-particle is double than that of proton. Reason: In a magnetic field, the time period of revolution of a charged particle is directly proportional to mass.

Join BYJU'S Learning Program

For periodic motion of small amplitude A, the time period T of this particle is proportional to

The correct option is A 1A√mαThe dimensions of α are V(x)x4=EnergyLength4=ML2T−2L4=[ML−2T−2]It is clear that 1A√mα has dimensions of time.

For periodic motion of small amplitude $A$ , the time period $T$ of this particle is proportional to

The correct option is A $\frac{1}{A} \sqrt{\frac{m}{α}}$
The dimensions of $α$ are $\frac{V (x)}{x^{4}} = \frac{E n e r g y}{{L e n g t h}^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$
It is clear that $\frac{1}{A} \sqrt{\frac{m}{α}}$ has dimensions of time.