The graph shows how the displacement of a particle performing simple harmonic motion varies with time - AQA - A-Level Physics - Question 32 - 2017 - Paper 1
Question 32
The graph shows how the displacement of a particle performing simple harmonic motion varies with time.
Which statement is not correct?
A. The speed of the particle... show full transcript
Worked Solution & Example Answer:The graph shows how the displacement of a particle performing simple harmonic motion varies with time - AQA - A-Level Physics - Question 32 - 2017 - Paper 1
Step 1
A. The speed of the particle is a maximum at time \( \frac{T}{4} \)
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
In simple harmonic motion (SHM), the speed of the particle is indeed at its maximum when it passes through the equilibrium position. Since ( \frac{T}{4} ) marks the time when the displacement is zero (the particle is at the equilibrium position), this statement is correct.
Step 2
B. The potential energy of the particle is zero at time \( \frac{3T}{4} \)
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
At time ( \frac{3T}{4} ), the particle is at its maximum displacement in the negative direction. Since potential energy is maximum at maximum displacement, this statement is not correct.
Step 3
C. The acceleration of the particle is a maximum at time \( \frac{T}{2} \)
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
At time ( \frac{T}{2} ), the particle is at maximum displacement in the positive direction, which corresponds to maximum acceleration. Therefore, this statement is correct.
Step 4
D. The restoring force acting on the particle is zero at time \( T \)
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
At time ( T ), the particle returns to the equilibrium position where the restoring force is also zero, confirming this statement is correct.
