Question
A SHM is expressed by the equation $\text{x}=\text{A}\cos(\omega\text{t}+\phi)$ and the phase angle $\phi=0$ . Draw graphs to show variation of displacement, velocity and acceleration for one complete cycle in SHM.
✓
Answer
Let $\text{x}=\text{A}\cos(\omega\text{t}+\phi)$ and if phase $\phi$ is zero, then $\text{x}=\text{A}\cos\omega\text{t}+\phi$
With the given data we plot x-t.v-t and a-t graphs. The grapha have beem shown in Fig. (a), (b) and (c).
$\therefore\text{v}=\frac{\text{dx}}{\text{dt}}$ $=-\text{A}\omega\sin\omega\text{t}$
$\text{a}\frac{\text{dv}}{\text{dt}}=-\text{A}\omega^2\cos\omega\text{t}$
$=-\omega\text{x}$
Thus, values of x ,v and a at different times, over one complete oscillation cycle are:| $\text{time}$ | $0$ | $\frac{\text{T}}{4}$ | $\frac{\text{T}}{2}$ | $\frac{3\text{T}}{4}$ | $\text{T}$ |
| $\omega\text{t}$ | $0$ | $\frac{\pi}{2}$ | $\pi$ | $\frac{3\pi}{2}$ | $2\pi$ |
| $\text{x}$ | $\text{A}$ | $0$ | $-\text{A}$ | $0$ | $\text{A}$ |
| $\text{v}$ | $0$ | $-\text{A}\omega$ | $0$ | $+\text{A}\omega$ | $0$ |
| $\text{a}$ | $-\text{A}\omega^2$ | $0$ | $\text{A}\omega^2$ | $0$ | $-\text{A}\omega^2$ |
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