Question
Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1 : 2 : 3 : 4.
✓
Answer
Let n be the number of loop in the string.The length of each loop is $\frac{\lambda}{2}$
$\therefore\text{L}=\frac{\text{n}\lambda}{2}$ or $\lambda=\frac{2\text{L}}{\text{n}}$
$\text{v}=\text{v}\lambda$ and $\lambda=\frac{\upsilon}{\text{v}}.$
so $\frac{\upsilon}{\text{v}}=\frac{2\text{L}}{\text{n}}$
$\text{v}=\frac{\text{n}}{2\text{L}}.\text{v}$ v is stretch string $=\sqrt{\frac{\text{T}}{\text{m}}}$
$\therefore\text{v}=\frac{\text{n}}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}$
For n =1, $\text{v}_1=\frac{1}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=\text{v}_0$
If n = 2 then $\text{v}_2=\frac{2}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=2\text{v}_0$
n = 3 then $\text{v}_3=\frac{3}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=3\text{v}_0$
$\therefore\text{v}_1:\text{v}_2:\text{v}_3:\text{v}_4:=\text{n}_1:\text{n}_2:\text{n}_3:\text{n}_4:=1:2:3:4$
$\therefore\text{L}=\frac{\text{n}\lambda}{2}$ or $\lambda=\frac{2\text{L}}{\text{n}}$
$\text{v}=\text{v}\lambda$ and $\lambda=\frac{\upsilon}{\text{v}}.$
so $\frac{\upsilon}{\text{v}}=\frac{2\text{L}}{\text{n}}$
$\text{v}=\frac{\text{n}}{2\text{L}}.\text{v}$ v is stretch string $=\sqrt{\frac{\text{T}}{\text{m}}}$
$\therefore\text{v}=\frac{\text{n}}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}$
For n =1, $\text{v}_1=\frac{1}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=\text{v}_0$
If n = 2 then $\text{v}_2=\frac{2}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=2\text{v}_0$
n = 3 then $\text{v}_3=\frac{3}{2\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}=3\text{v}_0$
$\therefore\text{v}_1:\text{v}_2:\text{v}_3:\text{v}_4:=\text{n}_1:\text{n}_2:\text{n}_3:\text{n}_4:=1:2:3:4$
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