Standing waves

Question

A stretched string of length 1m and mass 5 × ${10}^{\u20134}$ kg, fixed at both ends, is under a tension of 20 N. If it is plucked at points situated at 25 cm from one end, it would vibrate with a frequency :

Moderate

Solution

Standing waves of produced in the string attached on both the ends, whose frequency is given by $\frac{\mathrm{n}}{2\mathrm{l}}\sqrt{\frac{\mathrm{T}}{\mathrm{\mu}}}$

$\begin{array}{l}\mathrm{T}=20\mathrm{N}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\mu}=5\times {10}^{-4}\\ \therefore \mathrm{frequency}=\frac{\mathrm{n}}{2}\sqrt{4\times {10}^{4}}=100\text{\hspace{0.17em}}\mathrm{n}\end{array}$

Now according to question the string is plucked at 25 cm

Thus antinode is present at 25 cm, thus $\frac{\mathrm{\lambda}}{4}=25\mathrm{cm}$

$\mathrm{\lambda}=100\mathrm{cm}$

$\therefore \mathrm{n}=2$, hence frequency = 200Hz

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