A longitudinal wave is represented by x=x_0 s | vedclass

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

Question

(C) The given wave equation is $x = x_0 \sin 2 \pi (nt - x/\lambda)$.Comparing this with the standard form $x = A \sin (\omega t - kx)$,we have amplit

Vedclass · Accepted Answer

$\lambda=\frac{\pi x_0}{2}$

Vedclass · Answer

(C) The given wave equation is $x = x_0 \sin 2 \pi (nt - x/\lambda)$.Comparing this with the standard form $x = A \sin (\omega t - kx)$,we have amplitude $A = x_0$,angular frequency $\omega = 2 \pi n$,and wave number $k = 2 \pi / \lambda$.The maximum particle velocity $V_p$ is given by $V_p = A \omega = x_0 (2 \pi n) = 2 \pi n x_0$.The wave velocity $V_w$ is given by $V_w = \frac{\omega}{k} = \frac{2 \pi n}{2 \pi / \lambda} = n \lambda$.According to the problem,$V_p = 4 V_w$.Substituting the expressions,we get $2 \pi n x_0 = 4 (n \lambda)$.Dividing both sides by $n$,we get $2 \pi x_0 = 4 \lambda$.Therefore,$\lambda = \frac{2 \pi x_0}{4} = \frac{\pi x_0}{2}$.

$A$ longitudinal wave is represented by $x=x_0 \sin 2 \pi(n t-x / \lambda)$. The maximum particle velocity will be four times the wave velocity if:

Similar Questions

Explain linearly polarized waves and provide their definition.

An equation of a simple harmonic progressive wave is given by $y=A \sin (100 \pi t-3 x)$. The distance between two particles having a phase difference of $\frac{\pi}{18}$ in meters is:

$A$ transverse wave is travelling along a stretched string from right to left. The figure shown represents the shape of the string at a given instant. At this instant,

The figure represents the instantaneous picture of a longitudinal harmonic wave travelling along the negative $x$-axis. Identify the correct statement$(s)$ related to the movement of the points shown in the figure. The stationary points are

The two waves represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ have a phase difference of

Vedclass Test Series

Exam Paper Generator

Online Exam Module