Question

A transverse wave is represented by the equation
$y=y_0\sin \frac{2\pi }{\lambda }(vt-x)$
For what value of $\lambda ,$ the maximum particle velocity is equal to two times the wave velocity?

• A. $\lambda =\pi y_0/3$
• B. $\lambda =\pi y_0/2$
• C. $\lambda =\pi y_0$
• D. $\lambda =2\pi y_0$

Answer 1

The correct option is C $\lambda =\pi y_0$

Given:
A transverse wave is represented by the equation
$y=y_0\sin \frac{2\pi }{\lambda }(vt-x)$

Hence, on comparing the given equation with standard equation
$y=a\sin \frac{2\pi }{\lambda }(vt-x)$.

As we know,
Speed of wave $(v{)}_{wave}=v$
Maximum particle velocity
$(v{)}_{particle}=a\omega =y_0\times \text{co-efficient of}t$
$=y_0\times \frac{2\pi v}{\lambda }$

Therefore,
$(v_{max}{)}_{particle}=2(v{)}_{wave}$
$\Rightarrow \frac{a\times 2\pi v}{\lambda }=2v$

From above equation,
$\Rightarrow \lambda =\pi y_0$

Final answer: $\left(d\right)$