Exams › JEE Advanced › Physics

A block of mass m1 is attached to a spring of spring constant K. The spring's other end is connected over a smooth (massless... wait, mass m2) pulley. Find the time period of small oscillations of mass m2 (the pulley of mass m2).

1. T = 2*pi * sqrt((m1 + m2)/K)
2. T = 2*pi * sqrt((m1 + 4*m2)/K)
3. T = 2*pi * sqrt((4*m1 + m2)/K)
4. T = 2*pi * sqrt((3*m1 + m2)/K)

Correct answer: T = 2*pi * sqrt((m1 + 4*m2)/K)

Solution

In the standard pulley-spring-mass system where a smooth pulley of mass m2 has the spring attached to its axle and mass m1 hangs from the string over it, the effective oscillating mass is (m1 + 4*m2), giving T = 2*pi*sqrt((m1+4*m2)/K).

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 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?

⚔️ Practice JEE Advanced Physics free + battle 1v1 →