© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the known parameters
• Amplitude, $A = 5\,\text{cm} = 0.05\,\text{m}$ (converting to meters if required in SI units).
• Displacement from the mean position, $x = 4\,\text{cm} = 0.04\,\text{m}$. (For simplicity in symbolic manipulation, you may directly use cm, but SI would be in meters.)
• Condition given: The magnitude of velocity equals the magnitude of acceleration at $x=4\,\text{cm}$.
Step 2: Write the velocity and acceleration formulas in SHM
The velocity of a particle executing simple harmonic motion (SHM) is:
$v = \omega \sqrt{A^2 - x^2}$
The acceleration of the particle in SHM is:
$a = -\,\omega^2 x
(The negative sign indicates direction, but we only need magnitudes here.)
Step 3: Apply the condition |v| = |a|
Using the condition that the magnitudes are equal, we get:
$\omega \sqrt{A^2 - x^2} = \omega^2 \, x
Assuming $\omega \neq 0$, we can divide both sides by $\omega$ to obtain:
$\sqrt{A^2 - x^2} = \omega \, x
Squaring both sides:
$A^2 - x^2 = \omega^2 \, x^2
Step 4: Substitute the given values and solve for $\omega$
Here, $A = 5\,\text{cm}$ and $x = 4\,\text{cm}$. Substituting these into the equation:
$25 - 16 = \omega^2 \times 16
$9 = 16\,\omega^2 \quad\Longrightarrow\quad \omega^2 = \frac{9}{16}
$\omega = \frac{3}{4}\,\text{rad s}^{-1}
Step 5: Find the time period $T$
The time period of SHM is given by:
$T = \frac{2\pi}{\omega}
Substitute $\omega = \frac{3}{4}$:
$T = \frac{2\pi}{\frac{3}{4}} = \frac{2\pi \times 4}{3} = \frac{8\pi}{3}\,\text{seconds}
Final Answer
The periodic time of the simple harmonic motion is
$ \frac{8\pi}{3} \,\text{seconds}$.
