Question

A plane progressive transverse wave at an instant is shown below.

Choose an correct option.

• A. $\text{K.E}$ associated with particle $A$ is maximum
• B. $\text{P.E}$ associated with particle $C$ is maximum
• C. $\text{P.E}$ associated with particle $B$ is maximum
• D. $\text{K.E}$ associated with particle $D$ is minimum

Answer 1

The correct option is B $\text{P.E}$ associated with particle $C$ is maximum

At extreme position, the particle has maximum displacement associated with them and velocity of particle is zero.

Here total energy of particle,
$\Rightarrow \text{T.E}=\text{P.E}+\text{K.E}$
If kinetic energy is minimum then potential energy will be maximum.

Also due to maximum vertical displacement of particle at the extreme position $(A,C)$ the potential energy of particle will be maximum.
$\therefore$ Option $\left(b\right)$ is correct.