A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change ?
We have, $\mu=\frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{V} \rho}{\mathrm{L}}=\frac{\mathrm{AL} \rho}{\mathrm{L}}=\mathrm{A} \rho$
$\therefore m=\left(\pi r^{2}\right) \rho$
$\ldots(1)$
Now, we have, $f=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$
$\therefore f=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\pi r^{2} \rho}} \Rightarrow f \propto \frac{1}{r}$
( $\because$ other factors are constants)
$\therefore \frac{f_{1}}{f_{2}}=\frac{r_{2}}{r_{1}}=\frac{3}{1} \quad\left(\because r_{2}=3 r_{1}\right)$
$\therefore f_{2}=\left(\frac{1}{3}\right) f_{1}$
The fundamental frequency of vibration of a string stretched between two rigid support is $50\,Hz$. The mass of the string is $18\,g$ and its linear mass density is $20\,g / m$. The speed of the transverse waves so produced in the string is $..........\,ms ^{-1}$
In an experiment with sonometer when a mass of $180\,g$ is attached to the string, it vibrates with fundamental frequency of $30\,Hz$. When a mass $m$ is attached, the string vibrates with fundamental frequency of $50\,Hz$. The value of $m$ is $.........\,g$.
A tuning fork of frequency $392 Hz,$ resonates with $50 cm $ length of a string under tension ($T$). If length of the string is decreased by $2\%$, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
Two wires are in unison. If the tension in one of the wires is increased by $2\%, 5$ beats are produced per second. The initial frequency of each wire is .... $Hz$
The equation of a standing wave in a string fixed at both ends is given as $y=2 A \sin k x \cos \omega t$ The amplitude and frequency of a particle vibrating at the mid of an antinode and a node are respectively