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Q. A piece of wire cut into two pieces A and B and stretched to the same tension and minuted between two rigid walls. Segment A is longer than segment B. Which of the following quantities will always be larger for waves on A than for waves on B

Waves

Solution:

frequency $f =\frac{1}{21} \sqrt{\frac{ t }{\mu}}$
$f =\frac{ v }{\lambda}$
$\Rightarrow \frac{ v }{\lambda}=\frac{1}{21} \sqrt{\frac{ t }{\mu}}$
$\Rightarrow \lambda=21$
$\lambda\, \alpha\, l $
$\Rightarrow $ wavelength of $A$ is larges than that of $B$