Question

Two simple harmonic motions of angular frequency $100{\mathrm{rads}}^{-1}$ and $1000{\mathrm{rads}}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is

• A. $1:10$
• B. $1:100$
• C. $1:1000$
• D. $1:10000$

Answer 1

The correct option is B

$1:100$

Step 1: Given data

$$\begin{gathered} {\omega }_1=100{\mathrm{rads}}^{-1} \\ {\omega }_2=1000{\mathrm{rads}}^{-1} \end{gathered}$$

Here,

${\omega }_1$ is the angular velocity for the first simple harmonic motion.

${\omega }_2$ is the angular velocity for the second simple harmonic motion.

Step 2: The maximum acceleration of simple harmonic motion

The equation for the maximum acceleration of a simple harmonic equation can be written as:

$a_{\max }=-{\omega }^{2}A$

Where, $a_{\max }$ is the maximum acceleration, $\omega$ is the angular velocity, and $A$ is the displacement amplitude.

Step 3: Equation for the maximum acceleration for the first simple harmonic motion

Consider the displacement amplitude for the first simple harmonic motion as $A$.

${\left({\mathrm{a}}_{\max }\right)}_1=-{\left(100{\mathrm{rads}}^{-1}\right)}^{2}A$

Step 4: Equation for the maximum acceleration for the second simple harmonic motion

Since both the simple harmonic equations have the same displacement amplitude, the amplitude for the second simple harmonic motion is also $A$.

${\left(a_{\max }\right)}_2=-{\left(1000{\mathrm{rads}}^{-1}\right)}^{2}A$

Step 5: Calculation of ratio of the maximum accelerations

$$\begin{gathered} \frac{{\left(a_{\max }\right)}_1}{{\left(a_{\max }\right)}_2}=\frac{{100}^2}{{1000}^2} \\ \frac{{\left(a_{\max }\right)}_1}{{\left(a_{\max }\right)}_2}=\frac{10000}{1000000} \\ {\left(a_{\max }\right)}_1:{\left(a_{\max }\right)}_2=1:100 \end{gathered}$$

Hence, the ratio of the maximum accelerations is $1:100$.

Therefore, option (B) is correct.