Total energy of the normal mode on a string

[LCSphysicist]

Homework Statement

All below

Relevant Equations

All below

What you think about this question?

Seems a little strange to me, that is, it considers the maximum kinetic energy when the displacement of the oscillators is maximum, i don't think this is right.

 

[LCSphysicist]

Anyway, we could solve it by:

Just take theta equal zero (it doesn't matter) and

(omega n is just one thing, i don't know how to write the little n)
BTW, this will implies y = 0, where AFAIK is the true position where v is maximum
...

 

[TSny]

Seems a little strange to me, that is, it considers the maximum kinetic energy when the displacement of the oscillators is maximum, i don't think this is right.

The max KE occurs when the velocity $\dot y_n$ is max. So, the first equality in the equation is correct. For SHM $(\dot y_n)_{\rm max}$ has a value equal to $\omega_n (y_n)_{\rm max}$. So, you get the second equality. The relation $(\dot y_n)_{\rm max} = \omega_n (y_n)_{\rm max}$ is not meant to imply that the maximum value of $\dot y_n$ occurs at the same instant of time as the maximum value of $y_n$.

 

[LCSphysicist]

The max KE occurs when the velocity $\dot y_n$ is max. So, the first equality in the equation is correct. For SHM $(\dot y_n)_{\rm max}$ has a value equal to $\omega_n (y_n)_{\rm max}$. So, you get the second equality. The relation $(\dot y_n)_{\rm max} = \omega_n (y_n)_{\rm max}$ is not meant to imply that the maximum value of $\dot y_n$ occurs at the same instant of time as the maximum value of $y_n$.

Yeh...You are right, thank you.