Question

The period of oscillations of a point is 0.04 sec. and the velocity of propagation of oscillation is 300 m/sec. The difference of phases between the oscillations of two points at distances 10 and 16m respectively from the source of oscillations is:

• A. $2\pi$
• B. $\pi /2$
• C. $\pi /4$
• D. $\pi$

Answer 1

Answer: (D)

$\pi$

$\text{Time Period of the wave is given as :}t=0.04\text{seconds}$
$\text{Velocity of propagation of oscillation :}v=300\text{m/second}$
$\text{Wavelength of the oscillation}=\lambda =v\times t$
$\text{Now substituting the values :}$
$\lambda =300\times 0.04\lambda =12m$
$\text{Now using the relation between path difference}△x\text{and phase difference}△\varphi$
$\frac{△x}{\lambda }=\frac{△\varphi }{2\pi }$
$\text{Now finding the phase difference at 10 m from the source of oscillation:}$
$△x=10$
$\frac{10}{12}=\frac{△\varphi }{2\pi }\frac{10}{12}\times 2\pi =△\varphi △\varphi =\frac{5\pi }{3}$
$\text{Now finding the phase difference at 16 m from the source of oscillation:}$
$△x=16$
$\frac{16}{12}=\frac{△\varphi }{2\pi }\frac{16}{12}\times 2\pi =△\varphi △\varphi =\frac{8\pi }{3}$
$\text{Now finding the phase difference at 16m and 10 m from the source:}$
${△\varphi }_{16}-{△\varphi }_{10}=\frac{8\pi }{3}-\frac{5\pi }{3}{△\varphi }_{16}-{△\varphi }_{10}=\frac{3\pi }{3}{△\varphi }_{16}-{△\varphi }_{10}=\pi$