A simple harmonic motion having an amplitude | vedclass

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Question

(A) The standard equation for simple harmonic motion is given by $y = A \sin(\omega t + \phi)$,where $\omega = \frac{2\pi}{T}$.Given equation: $y = 5

Vedclass · Accepted Answer

$A = 5; T = 2$

Vedclass · Answer

(A) The standard equation for simple harmonic motion is given by $y = A \sin(\omega t + \phi)$,where $\omega = \frac{2\pi}{T}$.Given equation: $y = 5 \sin(\pi t + 4\pi)$.Comparing this with the standard form $y = A \sin(\omega t + \phi)$:The amplitude $A = 5 \ m$.The angular frequency $\omega = \pi \ rad/s$.Since $\omega = \frac{2\pi}{T}$,we have $\pi = \frac{2\pi}{T}$.Solving for $T$,we get $T = \frac{2\pi}{\pi} = 2 \ s$.Therefore,$A = 5$ and $T = 2$.

$A$ simple harmonic motion having an amplitude $A$ and time period $T$ is represented by the equation: $y = 5 \sin \pi (t + 4) \ m$. Then the values of $A$ (in $m$) and $T$ (in $sec$) are:

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