If x=5 sin left(pi t+frac{pi}{3}right) text{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The general equation for simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.Comparing the given equation $x = 5 \sin \left(\pi t + \

Vedclass · Accepted Answer

$5 	ext{ m}, 2 	ext{ s}$

Vedclass · Answer

(A) The general equation for simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.Comparing the given equation $x = 5 \sin \left(\pi t + \frac{\pi}{3}ight) 	ext{ m}$ with the general equation:Amplitude $A = 5 	ext{ m}$.Angular frequency $\omega = \pi 	ext{ rad/s}$.The time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$: $T = \frac{2\pi}{\pi} = 2 	ext{ s}$.Thus,the amplitude is $5 	ext{ m}$ and the time period is $2 	ext{ s}$.

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

Similar Questions

$A$ particle free to move along the $x$-axis has potential energy given by $U(x) = k[1 - \exp(-x^2)]$ for $-\infty \le x \le +\infty$,where $k$ is a positive constant of appropriate dimensions. Then:

$A$ particle executing simple harmonic motion along the $Y$-axis has its motion described by the equation $y = A \sin(\omega t) + B$. The amplitude of the simple harmonic motion is

$A$ body is moving in a room with a velocity of $20 \ m/s$ perpendicular to the two walls separated by $5 \ m$. There is no friction and the collisions with the walls are elastic. The motion of the body is

Which of the following expressions represent simple harmonic motion?

Draw plots for initial phase $\phi = 0$ for different periods.

Vedclass Test Series

Exam Paper Generator

Online Exam Module