Physics - Wave on String Question with Solution | TestHub

For fundamental harmonics of a closed end pipe, $\frac{\lambda }{4}=L$ Velocity of sound is in air is given by, $v=\lambda f$ For second harmonics of a string, ${\lambda }^{\prime }=L^{\prime }$ the wavelength is the length of the string. Speed of sound on the wave is related as, $v_{\mathrm{S}}={\lambda }^{\prime }f$. Speed of sound is given by, $v_{\mathrm{S}}=\sqrt{\frac{T}{\mu }},T$ is the tension in the wire and $\mu$ the mass per unit length of the string. Since, the string in second harmonics is in resonance with the closed end pipe in fundamental mode, $\begin{array}{rl}& f=\frac{v}{4L}=\frac{v_{\mathrm{S}}}{L^{\prime }}=\frac{1}{L^{\prime }}\sqrt{\frac{T}{\mu }}\\ & \Rightarrow \frac{320\mathrm{m}/\mathrm{s}}{4\times 0.8\mathrm{m}}=\frac{1}{0.5\mathrm{m}}\sqrt{\frac{50\mathrm{N}}{\mu }}\\ & \Rightarrow \mu =\frac{1}{50}{\mathrm{kgm}}^{-1}\\ & \Rightarrow m=\mu L^{\prime }=\frac{0.5\mathrm{m}}{50{\mathrm{kg}}^{-1}\mathrm{m}}=10\mathrm{g}\end{array}$