Question 303: Oscillations

Question

The displacement of a particle executing S.H.M. is $$x=\mathrm{a} \sin (\omega t-\phi)$$. Velocity of the particle at time $$\mathrm{t}=\frac{\phi}{\omega}$$ is $$\left(\cos 0^{\circ}=1\right)$$

Accepted Answer

To calculate the velocity of a particle in simple harmonic motion (S.H.M.), we use the derivative of the displacement function with respect to time. The displacement function given is:
$$x = a \sin(\omega t - \phi)$$
The velocity $$v$$ is the first derivative of displacement $$x$$ with time $$t$$:
$$v = \frac{dx}{dt}$$
Let's compute the derivative:
$$v = \frac{d}{dt} \left[ a \sin(\omega t - \phi) \right]$$
$$v = a \cos(\omega t - \phi) \cdot \frac{d}{dt} (\omega t - \phi)$$
$$v = a \cos(\omega t - \phi) \cdot \omega$$
$$v = a \omega \cos(\omega t - \phi)$$
We need to calculate this velocity at time $$t = \frac{\phi}{\omega}$$:
$$v = a \omega \cos\left(\omega \cdot \frac{\phi}{\omega} - \phi \right)$$
$$v = a \omega \cos(\phi - \phi)$$
$$v = a \omega \cos(0^{\circ})$$
$$v = a \omega \cdot 1$$
$$v = a \omega$$
Therefore, the correct answer is Option B:
$$a \omega$$

Answer

$$\omega \cos \phi$$

Answer

$$\mathrm{a} \omega$$

Answer

$$\omega \cos 2 \phi$$

Answer

$$-\mathrm{a} \omega \cos 2 \phi$$

Oscillations

The displacement of a particle executing S.H.M. is $x=\mathrm{a} \sin (\omega t-\phi). Velocity of the particle at time \mathrm{t}=\frac{\phi}{\omega} is \left(\cos 0^{\circ}=1\right)

Choose the correct answer: