Question

A simple pendulum is displaced to the left of its equilibrium position and is released from rest. Set the origin of the coordinate system at the equilibrium position of the pendulum, and let counterclockwise be the positive angular direction. Assume air resistance is so small that it can be ignored.

Part A:

Beginning the instant the pendulum is released, select the graph that best matches the angular position vs. time graph for the pendulum.

Part B:
Beginning the instant the object is released; select the graph that best matches the velocity vs. time graph for the object.

Part C:
Beginning the instant the object is released; select the graph that best matches the acceleration vs. time graph for the object.

 

Answer 1

Concepts and reason

The concept required to solve this problem is simple harmonic motion.

Initially, use the given statements to identify the correct position vs. time graph for the object. Then, use the conditions given to identify the velocity vs. time graph for the object. Finally, use the conditions given and simple harmonic motion definition to identify the acceleration vs. time graph for the object.

Fundamentals

The simple harmonic motion is the periodic motion in which a particle oscillates about a mean or equilibrium point. The force that restores the objects position is directly proportional to its displacement from the mean position and in opposite direction to the displacement.

The displacement, velocity, and acceleration are periodic functions for a simple harmonic that is there patterns repeat over a specific time.

Sign convention used here is as follows:

The upward direction is positive and downward direction is negative.

(A)

The simple pendulum is displaced to the left of its equilibrium position and is released from rest. Initially, the pendulum is displaced to the left so the angular displacement is negative so it implies that the position versus time graph has a negative intercept. This can be view from the restoring force also.

${\rm{restoring force}} = - mg\sin \theta$

Substitute ma for restoring force.

$\begin{array}{c}\\ma = - mg\sin \theta \\\\a = - g\sin \theta \\\end{array}$

For small value of the angle $\theta$ , the acceleration is directly proportional to the displacement which shows the simple harmonic motion.

Refer graph A, B, E and G the intercept for these is zero, so these are not correct options.

Refer graph F and C the intercept is positive while it should be negative here so these are incorrect.

Refer graph D, the curve is straight line while displacement will decrease and will be a repeated pattern so D is incorrect.

Refer graph H, the curve has a negative intercept and is a periodic repetition of up and downs. The best matching graph is H, as here the graph has negative intercept and shows a periodic variation.

(B)

The simple pendulum is displaced to the left of its equilibrium position and released from rest. The velocity is zero so there is no intercept or curve starts from origin. The simple pendulum will move towards the right from the left because of that the velocity will increase and is positive. The object will stop after reaching a maximum positive displacement so the curve will again touch time axis. The object undergoes a simple harmonic motion so the velocity graph is a periodic plot.

Refer graph C, D, F and H the intercept for these is not zero, so these are not correct options.

Refer graph B and G the curve is going in negative direction while the velocity is first increasing in positive direction so these are incorrect.

Refer graph A, the curve is straight line while velocity will again be zero after a while and will be a repeated pattern so A is incorrect.

Refer graph E, the intercept is zero and is a periodic repetition of up and downs starting in positive direction. The best matching graph is E, as here the graph has no intercept and shows a periodic variation increasing in positive direction first.

(C)

The simple pendulum is displaced to the left of its equilibrium position and released from rest. The acceleration is maximum initially and in upward direction. The acceleration will be zero as the object reaches mean position from down and will point downwards as the object starts moving upwards. The graph of acceleration versus time has a positive intercept and is a periodic function decreasing to zero.

Refer graph A, B, E and G, the intercept for these is zero, so these are not correct options.

Refer graph D and H, the intercept is negative while it should be positive here so these are incorrect.

Refer graph C, the curve is straight line while acceleration will decrease to zero and will be a repeated pattern so C is incorrect.

Refer graph F, the curve has a positive intercept and is a periodic repetition of up and downs starting to decrease to zero first. The best matching graph is F, as here the graph has positive intercept and shows a periodic variation.

Ans: Part A

The graph that best matches the position vs. time graph for the object is H.

Part B

The graph that best matches the velocity vs. time graph for the object is E.

Part C

The graph that best matches the acceleration vs. time graph for the object is F.