The maximum velocity of a particle executing simple harmonic motion is \(v\).
in Physics
13 views
3 Votes
3 Votes

The maximum velocity of a particle executing simple harmonic motion is \(v\). If the amplitude is doubled and the time period of oscillation decreased to \(1 / 3\) of its original value, the maximum velocity becomes

(a) \(18 \mathrm{v}\)

(b) \(12 v\)

(c) \(6 \mathrm{v}\)

(d) \(3 v\)

in Physics
by
4.3k Points

1 Answer

2 Votes
2 Votes
 
Best Answer
The correct option of this question will be (C).

Solution —

If $A$ is the amplitude of simple harmonic motion, then

$v=A \omega$

But $\omega=\frac{2 \pi}{T}$ where $T$ is the time period of oscillation.

$\therefore \quad v=A \frac{2 \pi}{T}$

When the amplitude is doubled and the time period decreased to $\frac{T}{3}$, the maximum velocity becomes

$v^{\prime}=2 A\left(\frac{2 \pi}{T / 3}\right)=6\left(A \frac{2 \pi}{T}\right)=6 v$

(using (i))
by
4.3k Points

Related Questions

0 Votes
0 Votes
1 Answer 21 Views

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]

Connect us on Social Media