Question
Two strings of the same material and length under the same tension may vibrate with different fundamental frequency. Why?
✓
Answer
The frequency of vibration of string is
given by $\text{n}=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$
m (mass per unit length) $=\frac{\text{mass}}{\text{length}}=\frac{\text{volume}\times\text{density}}{\text{l}}$
$=\frac{\pi\text{r}^2\text{l}\times\rho}{\text{l}}=\pi\text{r}^2\rho$
$=\pi\Big(\frac{\text{D}}{2}\Big)^2\rho=\frac{\pi\text{D}^2}{4}\rho$
$\text{n}=\frac{1}{2\text{l}}\sqrt{\frac{4\text{T}}{\pi\text{D}^2\rho}}=\frac{1}{\text{Dl}}\sqrt{\frac{\text{T}}{\pi\rho}}$
$\therefore \text{n}\propto\frac{1}{\text{D}}$ (when l, T, $\rho$ are same)
or nD = constant or $n_1D_1 = n_2D_2$
Hence the two strings may vibrate with different frequencies when they have different diameters.
given by $\text{n}=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$
m (mass per unit length) $=\frac{\text{mass}}{\text{length}}=\frac{\text{volume}\times\text{density}}{\text{l}}$
$=\frac{\pi\text{r}^2\text{l}\times\rho}{\text{l}}=\pi\text{r}^2\rho$
$=\pi\Big(\frac{\text{D}}{2}\Big)^2\rho=\frac{\pi\text{D}^2}{4}\rho$
$\text{n}=\frac{1}{2\text{l}}\sqrt{\frac{4\text{T}}{\pi\text{D}^2\rho}}=\frac{1}{\text{Dl}}\sqrt{\frac{\text{T}}{\pi\rho}}$
$\therefore \text{n}\propto\frac{1}{\text{D}}$ (when l, T, $\rho$ are same)
or nD = constant or $n_1D_1 = n_2D_2$
Hence the two strings may vibrate with different frequencies when they have different diameters.
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