Question

In the arrangement shown in figure, a mass can be hung from a string (with linear mass density of $\mu =0.002kg{m}^{-1}$ ) that passes over a light pulley. The string is connected to a vibrator (of constant frequency $f$ ), and the length of the string between point $P$ and the pulley is $L=2.00m$ . When the mass is either $9.0kg$ or $25.0kg$ , standing waves are observed. $\left(g=9.8m{s}^{-2}\right).$ What is the frequency (in $Hz$ ) of the vibrator to the nearest integer?

Answer 1

Answer: 263

$f=\frac{P}{2L}\sqrt{\frac{mg}{\mu }}.........\left(i\right)$
$P\sqrt{9}=\left(P-1\right)\sqrt{25}........\left(ii\right)$
$\therefore P=2.5$
$\therefore P\sqrt{9}=\left(P-2\right)\sqrt{25}$
$\therefore P=5$
$\therefore f=\frac{5}{2\times 2}\sqrt{\frac{88.2}{0.002}}=262.5\approx 263Hz$