The displacement from mean position of a particle in SHM at 3 seconds is `sqrt(3)` /2 of the amplitude. Its time period will be :

Options

A. 18 sec.
B. `6 sqrt3 sec`.
C. `9 sec.`
D. `3 sqrt3 sec`.

Answer

To solve the problem, we need to determine the time period of the simple harmonic motion (SHM) given that the displacement at 3 seconds is \(\frac{\sqrt{3}}{2}\) of the amplitude.

**Step 1: Understand the displacement in SHM**
The displacement \(Y\) of a particle in SHM can be expressed as:
\[ Y = A \sin(\omega t) \]
where \(A\) is the amplitude and \(\omega\) is the angular frequency.

**Step 2: Substitute the given values**
According to the problem, at \(t = 3\) seconds, the displacement \(Y\) is given as:
\[ Y = \frac{\sqrt{3}}{2} A \]
Substituting this into the SHM equation, we have:
\[ \frac{\sqrt{3}}{2} A = A \sin(\omega \cdot 3) \]

**Step 3: Simplify the equation**
We can divide both sides of the equation by \(A\) (assuming \(A \neq 0\)):
\[ \frac{\sqrt{3}}{2} = \sin(\omega \cdot 3) \]

**Step 4: Find the angle corresponding to the sine value**
From trigonometry, we know that:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus, we can equate:
\[ \omega \cdot 3 = \frac{\pi}{3} \]

**Step 5: Solve for \(\omega\)**
Now, we can solve for \(\omega\):
\[ \omega = \frac{\pi}{3 \cdot 3} = \frac{\pi}{9} \]

**Step 6: Relate angular frequency to time period**
The angular frequency \(\omega\) is related to the time period \(T\) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting our value of \(\omega\):
\[ \frac{\pi}{9} = \frac{2\pi}{T} \]

**Step 7: Solve for the time period \(T\)**
Cross-multiplying gives:
\[ \pi T = 18\pi \]
Dividing both sides by \(\pi\):
\[ T = 18 \]

Thus, the time period \(T\) of the simple harmonic motion is \(18\) seconds.

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