Question

A body undergoing simple harmonic motion has a maximum acceleration of \( 8\pi \text{ m/s}^2 \) and maximum speed of \( 1.6 \text{ m/s} \). What is the time period \( T \)?

• A. $0.1$ seconds
• B. $0.2$ seconds
• C. $0.3$ seconds
• D. $0.4$ seconds
• E. $0.5$ seconds

Hint:

A useful shortcut for SHM problems: the ratio $\frac{a_{max}}{v_{max}}$ always gives you $\omega$. Once you have $\omega$, finding $T$ or $f$ is straightforward. This avoids the need to calculate the amplitude $A$ separately unless specifically asked.

Answer 1

The Correct Option is D

Concept: In Simple Harmonic Motion (SHM), the maximum velocity ($v_{max}$) and maximum acceleration ($a_{max}$) are related to the amplitude ($A$) and angular frequency ($\omega$) as follows:
• $v_{max} = A\omega$
• $a_{max} = A\omega^2$ The time period $T$ is related to the angular frequency by $T = \frac{2\pi}{\omega}$.

Step 1: {Find the angular frequency $\omega$.}
We can find $\omega$ by taking the ratio of maximum acceleration to maximum velocity: $$\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$$ Substituting the given values: $$\omega = \frac{8\pi}{1.6}$$

Step 2: {Simplify the expression for $\omega$.}
$$\omega = \frac{80\pi}{16} = 5\pi \text{ rad/s}$$

Step 3: {Calculate the time period T.}
Using the relationship between $T$ and $\omega$: $$T = \frac{2\pi}{\omega}$$ $$T = \frac{2\pi}{5\pi}$$

Step 4: {Perform the final division.}
$$T = \frac{2}{5} = 0.4 \text{ seconds}$$ The time period of the oscillation is $0.4$ seconds.