Rankers Physics

Uncategorized: Practice Problem & Solution

A body is executing simple harmonic motion. When the displacements from the mean position is $4 \text{ cm}$ and $5 \text{ cm}$ the corresponding velocities of the body is $10 \text{ cm/sec}$ and $8 \text{ cm/sec}$. Then the time period of the body is: (1991)
$2\pi \text{ sec}$
$\pi/2 \text{ sec}$
$\pi \text{ sec}$
$(3\pi/2) \text{ sec}$

Solution Explained:

To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:

Using $v^2 = \omega^2(A^2 - x^2)$, we get $100 = \omega^2(A^2 - 16)$ and $64 = \omega^2(A^2 - 25)$.\nDividing them yields $A^2 = 41$. Substituting back gives $\omega^2 = 100/25 = 4 \implies \omega = 2$.\nTime period $T = 2\pi/\omega = \pi \text{ sec}$.

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