Question

In each of the four situations of column-I a stretched string or an organ pipe is given along with the required data. In case of strings the tension in string is T = 102.4 N and the mass per unit length of string is 1 g/m. Speed of sound in air is 320 m/s. The length of the strings and pipes are 0.5 m. Neglect end corrections. The frequencies of resonance are given in column-II. Match each situation in column-I with the possible resonance frequencies given in Column-II

Answer 1

Length of both pipes and the string $L=0.5m$

Tension in the string $T=102.4N$

Mass per unit length of string $\mu =1g/m=0.001kg/m$

Speed of sound in air $v=320m/s$

Now velocity of waves in the string $v_s=\sqrt{\frac{T}{\mu }}$ $=\sqrt{\frac{102.4}{0.001}}=320m/s$

Frequency of waves in the string closed at both ends, ${\nu }_s=\frac{n{v}_s}{2L}$

where $n=1,2,3,.......$

${\nu }_s=\frac{n320}{2\times 0.5}=320n$

$⟹{\nu }_s=320Hzand640Hz$

Frequency of waves in the string closed at one end, ${\nu }_s^{\prime }=\frac{(2n-1)v_s}{4L}$

where $n=1,2,3,.......$

${\nu }_s^{\prime }=\frac{(2n-1)320}{4\times 0.5}=(2n-1)160$

$⟹{\nu }_s^{\prime }=160Hzand480Hz$ for $n=1and2$ respectively.

Frequency of waves in the pipe an open at both ends, $\nu =\frac{nv}{2L}$ where $n=1,2,3,.......$

$\nu =\frac{n320}{2\times 0.5}=320n$

$⟹\nu =320Hzand640Hz$

Frequency of waves in the pipe closed at one end, ${\nu }^{\prime }=\frac{(2n-1)v}{4L}$ where $n=1,2,3,.......$

${\nu }^{\prime }=\frac{(2n-1)320}{4\times 0.5}=(2n-1)160$

$⟹{\nu }^{\prime }=160Hzand480Hz$ for $n=1and2$ respectively.