Question

At a moment in a progressive wave, the phase of the particle executing S.H.M is $\dfrac{\pi }{3}$ . Than the place of the particle 15cm ahead and at time T/2 will be, if the wavelength is 60cm
$\begin{align}
  & a)zero \\
 & b)\dfrac{\pi }{2} \\
 & c)\dfrac{5\pi }{6} \\
 & d)\dfrac{2\pi }{3} \\
\end{align}$

Answer 1

Hint: It is given in the question that we have to determine the phase between two particles at the time T/2. The angular frequency for both the particles is the same as they are two different particles of the same S.H.M. Hence we will use the equation of phase of a particle and take the difference between them to determine the phase of the particle 15cm ahead.

Formula used:
$\varphi =\omega t-kx$

Complete step-by-step answer:
Let us say a particle executes simple harmonic motion. The phase of the particle is given by,
$\varphi =\omega t-kx$ where $\omega $ is the angular frequency of the particle executing S.H.M, t is the instant of time for which the phase of the particle is to be calculated, $k=\dfrac{2\pi }{\lambda }$ where $\lambda $ is the wavelength the particle and x is the displacement or the position of the particle with respect to the origin. It is given to us that the phase of the particle 15cm behind is given as $\dfrac{\pi }{3}$ . hence substituting this in the above equation we get,
$\dfrac{\pi }{3}=\omega t-k{{x}_{0}}...(1)$ .
Let the phase of the particle 15cm ahead be${{\varphi }_{1}}$ .Therefore its phase can be written as,
${{\varphi }_{1}}=\omega t-k{{x}_{1}}...(2)$
We are asked to calculate the phase of the particle 15cm ahead at the same instant of time i.e. T/2. Therefore after taking the difference between equation 1 and 2 we get,
$\begin{align}
  & {{\varphi }_{1}}-\dfrac{\pi }{3}=\left( \omega t-k{{x}_{1}} \right)-\left( \omega t-k{{x}_{0}} \right) \\
 & \Rightarrow {{\varphi }_{1}}-\dfrac{\pi }{3}=\omega \dfrac{T}{2}-\dfrac{2\pi }{\lambda }{{x}_{1}}-\omega \dfrac{T}{2}+\dfrac{2\pi }{\lambda }{{x}_{0}} \\
 & \Rightarrow {{\varphi }_{1}}-\dfrac{\pi }{3}=\dfrac{2\pi }{\lambda }({{x}_{1}}-{{x}_{0}}) \\
 & \Rightarrow {{\varphi }_{1}}-\dfrac{\pi }{3}=\dfrac{2\pi }{60}(15) \\
 & \Rightarrow {{\varphi }_{1}}-\dfrac{\pi }{3}=\dfrac{\pi }{2} \\
 & \Rightarrow {{\varphi }_{1}}=\dfrac{\pi }{2}+\dfrac{\pi }{3}=\dfrac{5\pi }{6} \\
\end{align}$
Hence the correct answer of the above question is option c.

So, the correct answer is “Option c”.

Note: It is given that the two particles are at two different points. Hence we used the above equation to determine the phase difference between them. But if we are asked calculate the phase difference $\Delta \varphi $ between the same particle at two different instant of time such that the difference in time $\Delta t$ is given by $\Delta \varphi =\dfrac{2\pi }{T}\Delta t$ .