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Step-by-Step Solution
Step 1: Recall the Maximum Speed in SHM
For a particle executing Simple Harmonic Motion (SHM) with amplitude a and angular frequency \omega , the maximum speed v_{\max} is given by:
v_{\max} = a \, \omega.
Step 2: Write the Expression for Speed at a General Displacement
The speed v of the particle at displacement x from the mean position in SHM is:
v = \omega \sqrt{a^2 - x^2}.
Step 3: Use the Given Condition (Speed is Half of the Maximum)
The problem states that the particle’s speed is half of its maximum speed. Therefore, we set:
\frac{v_{\max}}{2} = \omega \sqrt{a^2 - x^2}.
Since v_{\max} = a \, \omega , we have:
\frac{a \, \omega}{2} = \omega \sqrt{a^2 - x^2}.
Step 4: Simplify the Equation
Divide both sides by \omega (assuming \omega \neq 0 ):
\frac{a}{2} = \sqrt{a^2 - x^2}.
Square both sides to remove the square root:
\left(\frac{a}{2}\right)^2 = a^2 - x^2.
\frac{a^2}{4} = a^2 - x^2.
Step 5: Solve for the Displacement x
Rearrange to find x^2 :
x^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}.
Hence,
x = \frac{\sqrt{3}}{2} \, a.
Step 6: Match with the Given Displacement Form
The problem states that the displacement (when speed is half the maximum) is given by \frac{\sqrt{x} \, a}{2}. We have found the actual displacement to be \frac{\sqrt{3}}{2} a.
So we set:
\frac{\sqrt{x} \, a}{2} = \frac{\sqrt{3} \, a}{2} \quad \Longrightarrow \quad \sqrt{x} = \sqrt{3} \quad \Longrightarrow \quad x = 3.
Final Answer
The value of x is 3.
