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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
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B
π/2
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C
π/4
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D
π
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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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