A string of length l is fixed at both ends and is vibrating in second harmonic. The amplitude at anti-node is 5 mm. The amplitude of a particle at distance `l//8` from the fixed end is

Options

A. `5sqrt(2)`mm
B. `(5)/(sqrt(2))mm`
C. 5mm
D. `(10)/(sqrt(2))mm`

Answer

A

Explanation

`A'=2Asinkx`
For second, `2.(lamda)/(2)=l implies lamda=l`
At x=l/8
`A'=2Asin((2pi)/(lamda)*(lamda)/(8))=2A"sin"(pi)/(4)=2A(((1)/(sqrt(2)))=(5)/(sqrt(2))m m`