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
2π
B
π/2
C
π/4
D
π

Solution

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

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