Question 358: Waves

Question

A string is stretched between two rigid supports separated by $$75 \mathrm{~cm}$$. There are no resonant frequencies between $$420 \mathrm{~Hz}$$ and $$315 \mathrm{~Hz}$$. The lowest resonant frequency for the string is

Accepted Answer

As there is no resonant frequency between $$315 \mathrm{~Hz}$$ and $$420 \mathrm{~Hz}$$, let $$315 \mathrm{~Hz}$$ be $$\mathrm{n}^{	ext {th }}$$ overtone and $$420 \mathrm{~Hz}$$ be $$(\mathrm{n}+1)^{	ext {th }}$$ overtone.
Now, $$v=\frac{\mathrm{nv}}{2 l}$$ ..... (i)
$$	herefore \quad 315=\frac{\mathrm{nv}}{2 l} 	ext { and } 420=\frac{(\mathrm{n}+1) \mathrm{v}}{2 l}$$
Taking the ratio,
$$\begin{aligned}
& \frac{315}{420}=\frac{n}{n+1} \
	herefore \quad & 315 n+315=420 n \
	herefore \quad & n=3
\end{aligned}$$
The resonant frequency is $$
u_0=\frac{v}{2 l}$$
Therefore, from equation (i) we get,
$$
u_0=\frac{
u}{\mathrm{n}}=\frac{315}{3}=105 \mathrm{~Hz}$$

Answer

$$210 \mathrm{~Hz}$$

Answer

$$180 \mathrm{~Hz}$$

Answer

$$105 \mathrm{~Hz}$$

Answer

$$1050 \mathrm{~Hz}$$

Waves

A string is stretched between two rigid supports separated by $75 \mathrm{~cm}. There are no resonant frequencies between 420 \mathrm{~Hz} and 315 \mathrm{~Hz}$. The lowest resonant frequency for the string is

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