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

Byju's Answer

Standard XII

Physics

SHM expression

A particle is...

Question

A particle is executing S.H.M with the amplitude of 4 cm and period 12 s. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is T1. The time taken from this displaced of 2 cm to reach the extreme position is T2. T1/T2 will be.

2

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

1

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

1 / 2

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

1 / 3

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

Open in App

Solution

The correct option is A $2$
$\frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}} = \frac{T_{1}}{T_{2}} = \frac{4}{2}$
$A α \frac{1}{f} α T$ or, $\frac{I_{2}}{T_{2}} = 2$
or, $T_{2} = 2 s$ .

Suggest Corrections

Similar questions

Q. Amplitude of oscillation of a particle that executes S.H.M. is 2 cm. Its displacement from its mean position in a time equal to 1/6th of its time period is

Q. A particle executing S.H.M. of amplitude 4 cm and T=4 sec. The time taken by it to move from positive extreme position(+a) to half the amplitude is

Q. Assertion :In simple harmonic motion A is the amplitude of oscillation. If

t_{1}

be the time to reach the particle from mean position to

\frac{A}{\sqrt{2}}

and

t_{2}

the time to reach from

\frac{A}{\sqrt{2}}

to A, then

t_{1} = \frac{t_{2}}{\sqrt{2}}

Reason : Equation of motion for the particle starting from mean position is given by

x = A c o s (ω t)

and of the particle starting from extreme position is given by

A s i n (ω t)

Q. A particle executing

S . H . M

of amplitude

4 c m

and

T = 4 s e c

. The time taken by it move from positive extreme position to half the amplitude is:-

A particle executes simple harmonic motion between x = -A and x=+ A. The time taken for it to go from the mean position, O to $\frac{A}{2}$ is $T_{1}$ and to go from $\frac{A}{2}$ to A is $T_{2}$ . Then,

Join BYJU'S Learning Program

A particle is executing S.H.M with the amplitude of 4 cm and period 12 s. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is T1. The time taken from this displaced of 2 cm to reach the extreme position is T2. T1/T2 will be.

The correct option is A 2T1T2=A1A2=T1T2=42Aα1fαT or, I2T2=2 or, T2=2s.

The correct option is A $2$
$\frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}} = \frac{T_{1}}{T_{2}} = \frac{4}{2}$
$A α \frac{1}{f} α T$ or, $\frac{I_{2}}{T_{2}} = 2$
or, $T_{2} = 2 s$ .