Question

A particle executing simple harmonic motion (SHM) has a maximum velocity of $1~ms^{-1}$ and a maximum acceleration of $2\pi~ms^{-2}$. Its time period of vibration is

• A. 1 s
• B. 2 s
• C. $\pi$ s
• D. $2\pi$ s
• E. 0.5 s

Hint:

A handy shortcut for SHM is $T = 2\pi \frac{v_{max}}{a_{max}}$. Plugging the numbers in directly: $2\pi \times \frac{1}{2\pi} = 1$. This saves time in competitive exams!

Answer 1

The Correct Option is A

Concept:
Physics - Simple Harmonic Motion (SHM).
Step 1: State the formulas for maximum velocity and acceleration.
In SHM, let $A$ be the amplitude and $\omega$ be the angular frequency.

• Maximum velocity ($v_{max}$) = $A\omega = 1~ms^{-1}$
• Maximum acceleration ($a_{max}$) = $A\omega^2 = 2\pi~ms^{-2}$

Step 2: Solve for angular frequency ($\omega$).
Divide the maximum acceleration by the maximum velocity: $$ \frac{A\omega^2}{A\omega} = \frac{2\pi}{1} $$ $$ \omega = 2\pi \text{ rad/s} $$
Step 3: Relate angular frequency to time period (T).
The relationship between $\omega$ and $T$ is: $$ \omega = \frac{2\pi}{T} $$
Step 4: Calculate the time period.
Substitute $\omega = 2\pi$ into the equation: $$ 2\pi = \frac{2\pi}{T} $$ $$ T = 1 \text{ s} $$ The time period of vibration is 1 second.