Topic: Engineering Mechanics
Speed of particle executing simple harmonic motion with amplitude 'a' is half of the maximum speed. At that instant, displacement of the particle is
- A. a/2
- B. ($\sqrt{3}$/2) * a
- C. (2a)/$\sqrt{3}$
- D. 3*$\sqrt{2}$ * a
Answer: B. ($\sqrt{3}$/2) * a
Explanation: Displacement x = a*sin(omega*t). Velocity v = dx/dt = a*omega*cos(omega*t). Given v = a*omega/2, we have a*omega*cos(omega*t) = a*omega/2, meaning cos(omega*t) = 1/2. Thus, omega*t = pi/3. At this instant, displacement x = a*sin(pi/3) = a*$\sqrt{3}$/2.