A particle executing S.H.M of amplitude 4 cm and T = 4 sec. The time taken (in sec) by it to move from positive extreme position to half the amplitude is :

Correct answer is between '0.665,0.669'. Can you explain this answer?

Question

A particle executing S.H.M of amplitude 4 cm and T = 4 sec. The time taken (in sec) by it to move from positive extreme position to half the amplitude is :

Correct answer is between '0.665,0.669'. Can you explain this answer?

Answer

Given:
Amplitude (A) = 4 cm
Time period (T) = 4 sec

To find:
Time taken by the particle to move from positive extreme position to half the amplitude

Formula:
The equation for Simple Harmonic Motion (SHM) is given by:
y = A * sin (2πt / T)

Where:
y = displacement of the particle at any given time t
A = amplitude of the motion
t = time
T = time period

Explanation:
To find the time taken by the particle to move from the positive extreme position to half the amplitude, we need to find the time when the displacement is half the amplitude.

Let's consider the equation for SHM:
y = A * sin (2πt / T)

We know that the displacement at the positive extreme position is equal to the amplitude (A).
So, when y = A/2, we can solve the equation to find the corresponding time (t).

Calculation:
Given:
A = 4 cm
T = 4 sec

Substituting the values in the equation:
A/2 = 4/2 = 2 cm

y = A * sin (2πt / T)
2 = 4 * sin (2πt / 4)

Dividing by 4:
1/2 = sin (2πt / 4)

Taking inverse sine of both sides:
sin^(-1)(1/2) = 2πt / 4

Using the inverse sine function, we find that sin^(-1)(1/2) = π/6

π/6 = 2πt / 4

Simplifying the equation:
1/6 = t/4

Cross multiplying:
4 = 6t

Solving for t:
t = 4/6 = 2/3 ≈ 0.667 sec

Therefore, the time taken by the particle to move from the positive extreme position to half the amplitude is approximately 0.667 seconds.

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