Exams › NEET › Physics

A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin (ωt + π/6). After the elapse of what fraction of the time period the velocity of the point will be equal to half its maximum velocity?

1. T/8
2. T/6
3. T/3
4. T/12

Correct answer: T/6

Solution

The maximum velocity in SHM is given by vmax = aω. The velocity at any time is v = aω cos(ωt + π/6). Setting v = vmax/2, we get cos(ωt + π/6) = 1/2. Solving for ωt + π/6 = π/3 or 5π/3, and converting to time, we find the required time is T/6.

Related NEET Physics questions

• One oscillation completed by a vibrating body in one second is known as:
• A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of time t. Which of the following statements is true. Assume that the particle started at \( t=0 \) at its mean position
• The amplitude of a particle executing SHM about O is 10 cm. Then: This question has multiple correct options
• The displacement (in \( \mathrm{m} \) ) of a particle of mass 100 gram from its equilibrium position is given by \( \boldsymbol{y}=\mathbf{0 . 0 1} \sin 2 \pi(t+ \) 0.4). Select the correct option(s): This question has multiple correct options
• Three simple harmonic motions in the same direction having the same amplitude \( A \) and same period are superposed. If each differs in phase from the next by \( 45^{0}, \) then
• A particle executes \( \boldsymbol{S} \boldsymbol{H} \boldsymbol{M} \) of amplitude \( 25 c m \) and time period \( 3 s \) What is the minimum time required for the particle to move between two points \( 12.5 \mathrm{cm} \) on either side of the mean position?

⚔️ Practice NEET Physics free + battle 1v1 →