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(B) In simple harmonic motion $(SHM)$, the maximum velocity is given by $v_{	ext{max}} = \omega a$, where $\omega$ is the angular frequency and $a$ i

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) In simple harmonic motion $(SHM)$, the maximum velocity is given by $v_{	ext{max}} = \omega a$, where $\omega$ is the angular frequency and $a$ is the amplitude.Given, $v_{	ext{max}} = 6.28 	ext{ cm s}^{-1}$.The length of the path is the total distance covered by the particle in one oscillation, which is equal to $2a$.So, $2a = 8 	ext{ cm}$, which implies $a = 4 	ext{ cm}$.Using the formula $v_{	ext{max}} = \omega a$, we have $\omega = \frac{v_{	ext{max}}}{a} = \frac{6.28}{4} 	ext{ rad s}^{-1}$.Since $\omega = \frac{2\pi}{T}$, we can write $\frac{2\pi}{T} = \frac{6.28}{4}$.Substituting $\pi \approx 3.14$, we get $\frac{2 	imes 3.14}{T} = \frac{6.28}{4}$.$\frac{6.28}{T} = \frac{6.28}{4}$.Therefore, $T = 4 	ext{ s}$.

The maximum velocity of a particle performing simple harmonic motion is $6.28 \text{ cm s}^{-1}$. If the length of its path is $8 \text{ cm}$, then what is its period (in $\text{ s}$)?

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