The velocity time graph of a particle executing SHM is shown in figure...
Velocity-Time Graph of a Particle executing SHM:
The given velocity-time graph represents the motion of a particle executing Simple Harmonic Motion (SHM). We need to analyze the graph to determine the correct statement among the given options.
a) Force is maximum at t = (3T)/4:
- The magnitude of force acting on a particle undergoing SHM is given by the equation F = -kx, where k is the spring constant and x is the displacement from equilibrium.
- From the graph, we observe that the velocity of the particle is maximum at t = (3T)/4. This corresponds to the maximum displacement of the particle from equilibrium.
- At maximum displacement, the force acting on the particle is maximum. Therefore, the statement "force is maximum at t = (3T)/4" is correct.
b) Potential energy of the particle is maximum at t = T/4:
- The potential energy of a particle undergoing SHM is given by the equation PE = (1/2)kx^2, where k is the spring constant and x is the displacement from equilibrium.
- From the graph, we observe that the velocity of the particle is zero at t = T/4. This corresponds to the particle being at equilibrium.
- At equilibrium, the displacement is maximum, and thus, the potential energy is maximum. Therefore, the statement "potential energy of the particle is maximum at t = T/4" is incorrect.
c) Kinetic energy of the particle is maximum at t = T:
- The kinetic energy of a particle undergoing SHM is given by the equation KE = (1/2)mv^2, where m is the mass of the particle and v is the velocity.
- From the graph, we observe that the velocity of the particle is maximum at t = T. This corresponds to the maximum displacement from equilibrium.
- At maximum displacement, the velocity is maximum, and thus, the kinetic energy is maximum. Therefore, the statement "kinetic energy of the particle is maximum at t = T" is correct.
d) Acceleration of the particle is maximum at t = T/4:
- The acceleration of a particle undergoing SHM is given by the equation a = -ω^2x, where ω is the angular frequency and x is the displacement from equilibrium.
- From the graph, we observe that the velocity of the particle is zero at t = T/4. This corresponds to the particle being at equilibrium.
- At equilibrium, the displacement is maximum, and thus, the acceleration is maximum. Therefore, the statement "acceleration of the particle is maximum at t = T/4" is correct.
In conclusion, the correct statements are:
a) Force is maximum at t = (3T)/4
c) Kinetic energy of the particle is maximum at t = T
d) Acceleration of the particle is maximum at t = T/4
The velocity time graph of a particle executing SHM is shown in figure...
Attach figure bro...
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