QJEEM24Phbb71: JEE Main 2024 Physics - Class 11 JEEM24Phbb71 Solution: The displacement of a particle executing SHM is given by . The…

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:

$\omega = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$

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:

$v = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$

Thus, the velocity of the particle at $t = 0$ is $10$ m/s.