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Question: In a string the speed of the wave is 10 \( m{s^{ - 1}} \) and its frequency is \( 100\,Hz \) . The v...

In a string the speed of the wave is 10 ms−1m{s^{ - 1}} and its frequency is 100 Hz100\,Hz . The value of the phase difference at a distance 2.5 cm2.5\,cm will be:
A.  π2 B.  π8 C.  3π2 D.  4π  A.\,\,\dfrac{\pi }{2} \\\ B.\,\,\dfrac{\pi }{8} \\\ C.\,\,\dfrac{{3\pi }}{2} \\\ D.\,\,4\pi \\\

Explanation

Solution

Hint
In the question, velocity and the frequency of the wave is given, substitute the values in the equation of the wavelength we get the value of wavelength and substituting all the known values in the equation of phase difference we get the value of phase difference at a required distance has been calculated.
The expression for finding the wavelength is
λ=vf\lambda = \dfrac{v}{f}
Where, λ \lambda \, be the wavelength of the wave, vv be the speed of the wave and ff be the frequency of the wave.
The expression for finding the phase difference is
ϕ=2πλ×x\phi = \dfrac{{2\pi }}{\lambda } \times x
Where xx be the distance of the phase difference.

Complete step by step answer
From the question, Given that
Speed of the wave v=10 ms−1v = 10\,m{s^{ - 1}}
Frequency of the wave f=100 Hzf = 100\,Hz
λ=vf.........(1)\lambda = \dfrac{v}{f}.........\left( 1 \right)
Substitute the value of vv and ff in the equation (1)\left( 1 \right)
λ=10100\lambda = \dfrac{{10}}{{100}}
Simplify the above equation, we get
λ=0.1 m\lambda = 0.1\,m
Therefore, the wavelength of the wave, λ=0.1 m\lambda = 0.1\,m
Now, we can find the value of phase difference at 2.5 cm2.5\,cm
Phase difference ϕ=2πλ×x.........(2)\phi = \dfrac{{2\pi }}{\lambda } \times x.........\left( 2 \right)
Here x=2.5 cmx = 2.5\,cm
Convert the xx value in terms of m,m, we get
x=2.5×10−2 mx = 2.5 \times {10^{ - 2}}\,m
Substitute the value of xx and λ\lambda in the equation (2)\left( 2 \right)
ϕ=2π0.1×(2.5×10−2)\phi = \dfrac{{2\pi }}{{0.1}} \times \left( {2.5 \times {{10}^{ - 2}}} \right)
Perform the arithmetic operations in the above equation, we get
ϕ=π2\phi = \dfrac{\pi }{2}
Therefore, the value of the phase difference at a distance of 2.5 cm2.5\,cm will be π2\dfrac{\pi }{2} .
Hence, from the above options, the option (A)\left( A \right) is correct.

Note
In the question, we know that the distance of the phase difference depends on the distance. If the distance value is increased or decreased the value of the phase difference is changed and given the value of the distance is in terms of cmcm but the value of wavelength is in terms of mm . So, we convert the distance in terms of m.m.