First slide

Simple harmonic motion

Question

The variation of velocity of a particle executing SHM with time is shown in figure. The velocity of the particle when a phase change of $π / 6$ takes place from the instant it is at one of the extreme positions will be

Moderate

A

3.53 m/s
B

2.5 m/s
C

4.330 m/s
D

None of these

Solution

From the graph T = (5 -1) = 4 s
(distance between the two adjacent crests shown in the figure)
$\begin{array}{l} and v_{max} = 5 m / s; ω A = 5 m / s \\ (\frac{2 π}{T}) A = 5 \Rightarrow A = \frac{5 T}{2 π} = \frac{5 \times 4}{2 π} = \frac{10}{π} m \\ Also, ω = \frac{2 π}{4} = \frac{π}{2} rad / s \end{array}$
The equation of velocity can be written as
$v = 5 \sin (\frac{π}{2} t) m / s$
$At extreme position, v = 0; \sin (\frac{π}{2} t) = 0 or t = 2 s$
Phase of the particle velocity at that instant corresponding
to the above equation = $π$ .
Therefore, when a phase change of $π$ /6 takes place, the
resulting phase = $π$ + $π$ /6.
$\begin{matrix} v = 5 \sin (π + \frac{π}{6}) = - 5 \sin \frac{π}{6} = - 5 (\frac{1}{2}) \\ = 2.5 m / s (numerically) \end{matrix}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App