Question 163: Oscillations

Question

The displacement of a particle executing SHM is given by $$x=10 \sin \left(w t+\frac{\pi}{3}\right) m$$. The time period of motion is $$3.14 \mathrm{~s}$$. The velocity of the particle at $$t=0$$ is _______ $$\mathrm{m} / \mathrm{s}$$.

Accepted Answer

The displacement of a particle executing Simple Harmonic Motion (SHM) can be expressed as:
$$x = A \sin(\omega t + \phi)$$
Where:
$$A$$ is the amplitude of the SHM,
$$\omega$$ is the angular frequency,
$$t$$ is the time,
$$\phi$$ is the phase constant (phase angle at $$t = 0$$).
In the given equation, $$x = 10 \sin(\omega t + \frac{\pi}{3})$$ m, the amplitude $$A = 10$$ m and the phase constant $$\phi = \frac{\pi}{3}$$. The time period $$T = 3.14$$ s is given, from which we can find the angular frequency $$\omega$$ using the relationship:
$$\omega = \frac{2\pi}{T}$$
Substituting the given $$T = 3.14$$ s:
$$\omega = \frac{2\pi}{3.14} \approx 2 \, 	ext{rad/s}$$
To find the velocity of the particle, we differentiate the displacement $$x$$ with respect to time $$t$$. The derivative of the displacement gives the velocity:
$$v = \frac{dx}{dt}$$
So, for $$x = 10 \sin(\omega t + \frac{\pi}{3})$$:
$$v = \frac{d}{dt}[10 \sin(\omega t + \frac{\pi}{3})]$$
Applying differentiation, we get:
$$v = 10\omega \cos(\omega t + \frac{\pi}{3})$$
Plug in the value of $$\omega = 2$$ rad/s and evaluate it at $$t = 0$$ to find the initial velocity:
$$v = 10 \cdot 2 \cos(2 \cdot 0 + \frac{\pi}{3})$$
$$v = 20 \cos(\frac{\pi}{3})$$
$$\cos(\frac{\pi}{3}) = \frac{1}{2}$$, therefore:
$$v = 20 \cdot \frac{1}{2} = 10 \, 	ext{m/s}$$
Thus, the velocity of the particle at $$t = 0$$ is $$10$$ m/s.

Oscillations

The displacement of a particle executing SHM is given by $x=10 \sin \left(w t+\frac{\pi}{3}\right) m. The time period of motion is 3.14 \mathrm{~s}. The velocity of the particle at t=0 is _______ \mathrm{m} / \mathrm{s}$.

Integer / Numeric Answer Type