Topic: Theory of Machines / Vibrations
Speed of particle executing simple harmonic motion with amplitude a is half of the maximum speed. At the instant, displacement of the particle is-
- A. $\frac{a}{2}$
- B. $\frac{\sqrt{3}}{2}a$
- C. $\frac{2a}{\sqrt{3}}$
- D. $3\sqrt{2}a$
Answer: B. $\frac{\sqrt{3}}{2}a$
Explanation: Equation for simple harmonic motion is $x = a \sin \omega t$. Velocity $v = \frac{dx}{dt} = a \omega \cos \omega t$. For maximum speed, $\cos \omega t = 1$, then $V_{max} = a\omega$. When speed is half of maximum, $\frac{a\omega}{2} = a\omega \cos \omega t \implies \cos \omega t = 1/2 \implies \omega t = \pi/3$. At $\omega t = \pi/3$, displacement $x = a \sin(\pi/3) = \frac{\sqrt{3}}{2}a$.