Question

A wave of frequency 500 Hz has a phase velocity of 360 m/s. The phase difference between the two displacements at a certain point in a time interval of 10${}^{-3}$ seconds will be how much?

• A. $\frac{\pi }{2}$ radian
• B. $\pi$ radian
• C. $\frac{\pi }{4}$ radian
• D. $\frac{\pi }{8}$ radian

Answer 1

Answer: (B)

$\pi$ radian

According to the concept of simple harmonic wave,

$y=asin\frac{2\pi }{\lambda }(vt-\varphi )$

Now, phase angle at point "$x$" for time"$t_1$"

$t_1=\frac{2\pi }{\lambda }(vt-x)$

now for time "$t_2$"

$t_2=\frac{2\pi }{\lambda }(vt-x)$

phase difference ,$({\varphi }_2-{\varphi }_1)=\frac{2\pi }{\lambda }(v{t}_2-x)-\frac{2\pi }{\lambda }(v{t}_1-x)$

$=\frac{2\pi }{\lambda }v(t_2-t_1)$

$=\frac{2\pi n\lambda }{\lambda }(t_2-t_1)$

$=2\pi n(t_2-t_1)$

$=2\pi \times 500\times {10}^{-3}$

$=\pi radian$