Standing Waves On A String

Question

A sonometer wire of length 0.6 m is fixed rigidly at both ends, and waves travel along it at 180 m per second. What is the lowest frequency at which this wire can vibrate in a standing wave pattern?

RankGuru · Accepted Answer

A string fixed at both ends supports standing waves with nodes at each end, and the fundamental mode fits exactly half a wavelength into the length, so $\lambda_1 = 2L$. This is the longest possible wavelength the boundary conditions permit and therefore corresponds to the lowest frequency the wire can sound. The fundamental frequency is $f_1 = v/\lambda_1 = v/(2L)$. Substituting $v = 180	ext{ m/s}$ and $L = 0.6	ext{ m}$ gives $f_1 = \frac{180}{2 	imes 0.6} = \frac{180}{1.2} = 150	ext{ Hz}$. The value $300	ext{ Hz}$ is actually the second harmonic, using $\lambda = L$ instead of $2L$. The value $75	ext{ Hz}$ wrongly uses $\lambda = 4L$, which applies to a pipe closed at one end, not a string fixed at both ends. The value $108	ext{ Hz}$ has no consistent harmonic basis. This follows the NCERT treatment of normal modes on a string with both ends fixed. As a plausibility check, the fundamental should be the smallest of the allowed harmonic series $f_n = n f_1$, and $150	ext{ Hz}$ correctly serves as the base from which higher harmonics like 300 Hz and 450 Hz are built.

Standing Waves On A String

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