Question

A body executing S.H.M. has a maximum velocity of $1 \text{ ms}^{-1}$ and a maximum acceleration of $4 \text{ ms}^{-2}$. Its amplitude in metres is:

• A. $1$
• B. $0.75$
• C. $0.5$
• D. $0.25$

Hint:

A useful combined formula to remember for these problems is $A = \frac{(v_{max})^2}{a_{max}}$. Plugging in the numbers directly: $1^2 / 4 = 0.25$. This saves time by skipping the explicit calculation of $\omega$.

Answer 1

The Correct Option is D

To find the amplitude of a body in Simple Harmonic Motion (S.H.M.), we utilize the relationships between maximum velocity, maximum acceleration, and angular frequency. Step 1: Identify the Formulas for S.H.M. For a body with amplitude $A$ and angular frequency $\omega$:

• Maximum Velocity ($v_{max}$) = $A\omega$

• Maximum Acceleration ($a_{max}$) = $A\omega^2$

Step 2: Calculate Angular Frequency ($\omega$) We can find $\omega$ by taking the ratio of maximum acceleration to maximum velocity: $$\omega = \frac{a_{max}}{v_{max}}$$ Substituting the given values ($v_{max} = 1 \text{ ms}^{-1}$ and $a_{max} = 4 \text{ ms}^{-2}$): $$\omega = \frac{4}{1} = 4 \text{ rad/s}$$

Step 3: Calculate Amplitude ($A$) Now, use the maximum velocity formula to solve for $A$: $$v_{max} = A\omega \implies A = \frac{v_{max}}{\omega}$$ $$A = \frac{1}{4} = 0.25 \text{ m}$$ Therefore, the amplitude of the motion is $0.25$ metres.