Question

If the equation for a transverse wave is $\mathrm{y}=\mathrm{A} \operatorname{Sin} 2 \pi\( \)\\{(1 / \mathrm{T})-(\mathrm{x} / \lambda)\\}$. then for what wavelength will the maximum velocity of the particle be double the wave velocity ? (A) \((\pi \mathrm{A} / 4)\) (B) \((\pi \mathrm{A} / 2)\) (C) \(\pi \mathrm{A}\) (D) \(2 \pi \mathrm{A}\)

Short answer

The short answer is: \(\lambda = \pi A\). The correct option is (C).

Step-by-step solution

Wave Equation

The given wave equation is: \[y = A\, \sin{(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right))}\]

Find maximum velocity

To find the maximum velocity of the particle, we need to differentiate the wave equation with respect to time. \[v_\text{particle}(t) = \frac{dy}{dt}\] Taking the partial derivative with respect to time, we get: \[v_\text{particle}(t) = 2\pi A \frac{1}{T} \cos{(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right))}\] To find the maximum velocity, we need to find when v_particle(t) reaches its maximum amplitude. This occurs when the value inside the cosine function is: \[\cos{(2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right))} = \pm 1\] Then, the maximum velocity will be: \[v_\text{max} = \pm 2\pi A \frac{1}{T}\]

Find the wave speed

The wave speed can be found using the following relation: \[v_\text{wave} = \frac{\lambda}{T}\]

Set the maximum velocity equal to twice the wave speed

Now, we can set the maximum velocity equal to twice the wave speed: \[v_\text{max} = 2 v_\text{wave}\] Substitute the expressions for v_max and v_wave: \[\pm 2\pi A \frac{1}{T} = 2\left(\frac{\lambda}{T}\right)\]

Solve for the wavelength

Now we can solve for the wavelength: \[\lambda = \pm \pi A\] Since we cannot have a negative wavelength, we'll take only the positive solution, so the answer is: \[\lambda = \pi A\] From the given options, the correct answer is (C) \(\pi \mathrm{A}\).