If $x = 5 \sin(\pi t + \frac{\pi}{3}) ext{ m}$ represents the motion of a particle executing simple harmonic

Question

If $x = 5 \sin(\pi t + \frac{\pi}{3}) 	ext{ m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively are:

Accepted Answer

1.  **Standard Equation of SHM:** The general displacement equation for a particle executing simple harmonic motion is given by $x(t) = A \sin(\omega t + \phi)$, where:
    *   $A$ is the **Amplitude** (maximum displacement from the mean position).
    *   $\omega$ is the **Angular Frequency**.
    *   $\phi$ is the **Phase Constant** (or initial phase).
2.  **Comparing Equations:** We compare the given equation $x = 5 \sin(\pi t + \frac{\pi}{3})$ with the standard form.
    *   The coefficient of the sine function corresponds to the amplitude: $A = 5$. Since the displacement $x$ is given in meters (m), the amplitude is $A = 5 	ext{ m}$.
    *   The coefficient of time $t$ inside the sine argument corresponds to the angular frequency: $\omega = \pi 	ext{ rad/s}$.
3.  **Calculate Time Period ($T$):** The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.
    *   Substituting the value of $\omega$: $T = \frac{2\pi}{\pi} = 2 	ext{ s}$.
4.  **Conclusion:** The amplitude is $5 	ext{ m}$ and the time period is $2 	ext{ s}$.

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