Exams › JEE Advanced › Physics

A particle undergoes simple harmonic motion with time period T. At a certain instant its speed is 60% of its maximum speed and is increasing. A time interval deltaₜ later, its speed becomes 80% of its maximum speed and is now decreasing. What is the smallest possible value of deltaₜ?

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

Correct answer: 3T/8

Solution

Writing v = v_max * sin(omega*t), speed is 60% of max at phase phi1 = arcsin(0.6) in the increasing portion (first quadrant), and 80% of max at phase phi2 = pi - arcsin(0.8) in the decreasing portion (second quadrant). The time difference is (phi2 - phi1)/omega, giving deltaₜ = (pi - arcsin(0.8) - arcsin(0.6)) / omega = (pi - 53.13 deg - 36.87 deg)*(T/2pi) = (180 - 90) deg * T/360 deg = 90 deg * T/360 = T/4... re-examining with the constraint that minimum deltaₜ spans from the 60% increasing phase to the 80% decreasing phase going forward in time.

Related JEE Advanced Physics questions

• A particle constrained to move along the x-axis has a potential energy described by U(x) = k[1 - exp(-x²)], where k is a positive constant and -∞ ≤ x ≤ +∞. Which of the following statements is true?
• What is the increase in potential energy of a spring when it is stretched or compressed by a distance x from its equilibrium position?
• A particle executes SHM of amplitude A and angular frequency omega along a straight line. When it is at a distance (sqrt(3)/2)A from the mean position, an impulsive force instantaneously increases its kinetic energy by (1/2)m*omega²*A². What is the new amplitude of oscillation?
• Two particles P and Q each perform SHM along the same straight line. Their displacements are given by y_P = A sin(omega*t + phi₁) and y_Q = A cos(omega*t + phi₂). At t = 0, particle P is at displacement A/sqrt(2) from mean position and moving toward positive displacement (rightward), while particle Q is at displacement (sqrt(3)/2)*A and moving toward negative displacement (leftward). What is the value of (phi₂ - phi₁)?
• A disc performs angular (torsional) oscillations using a wire with torsional constant k. Its initial angular amplitude is theta0. At the instant when the disc passes through angular displacement theta0/2 (half the amplitude), an impulsive torque is applied that instantaneously increases its angular velocity to 1.5 times its value at that position. Find the new angular amplitude of oscillation after the impulse.
• A block of mass m executes SHM of amplitude A on a frictionless surface. At the instant it is at the negative extreme position, a bullet of mass m moving in the negative x-direction with speed v0 strikes it and gets embedded. The condition mv0² = 2kA² holds, where k is the spring constant. The displacement x as a function of time t after the collision is:

⚔️ Practice JEE Advanced Physics free + battle 1v1 →