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What is the maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 3 mm and frequency of 20 Hz?
  • a)
    47 m/s2
  • b)
    48 m/s2
  • c)
    49 m/s2
  • d)
    50 m/s2
Correct answer is option 'A'. Can you explain this answer?
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Maximum Acceleration in Simple Harmonic Motion

In simple harmonic motion (SHM), the acceleration of a particle can be calculated using the equation:

a = -ω^2x

where a is the acceleration, ω is the angular frequency, and x is the displacement from the equilibrium position.

Given Data:
Amplitude (A) = 3 mm = 0.003 m
Frequency (f) = 20 Hz

Calculating Angular Frequency:
The angular frequency (ω) can be calculated using the formula:

ω = 2πf

Substituting the given frequency into the formula:

ω = 2π * 20 = 40π rad/s

Calculating Maximum Acceleration:
The maximum acceleration (amax) occurs when the particle is at maximum displacement, which is equal to the amplitude (A).

Substituting the values into the acceleration equation:

amax = -ω^2A

amax = -(40π)^2 * 0.003

amax = -1600π^2 * 0.003

amax ≈ -150.796 m/s^2

Since the acceleration is a vector quantity, the negative sign indicates that the acceleration is in the opposite direction to the displacement.

Converting to Positive Value:
To find the maximum positive acceleration, we can take the magnitude of the acceleration by removing the negative sign:

|amax| = 150.796 m/s^2

Rounding the value to the nearest whole number:

|amax| ≈ 151 m/s^2

Choosing the Correct Option:
Among the given options (a) 47 m/s^2, (b) 48 m/s^2, (c) 49 m/s^2, and (d) 50 m/s^2, the closest option to the calculated value of 151 m/s^2 is (a) 47 m/s^2.

Therefore, the correct answer is option 'A' - 47 m/s^2.
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