Waves: Tricky Questions Solved (5)

$v_1 = \sqrt{100/0.01} = 100$ m/s, $v_2 = \sqrt{144/0.01} = 120$ m/s.

For a string fixed at both ends, $f_n = n v / (2L)$. So $f_1^{(1)} = 100/2 = 50$ Hz, $f_1^{(2)} = 120/2 = 60$ Hz. Difference $= 10$ Hz.

We need $n_1 f_1^{(1)} = n_2 f_1^{(2)}$, i.e. $50 n_1 = 60 n_2$, or $5 n_1 = 6 n_2$. Smallest integers: $n_1 = 6, n_2 = 5$.

$f = 6 \times 50 = 300$ Hz, also $= 5 \times 60 = 300$ Hz. Confirmed.

The 6th harmonic of string 1 matches the 5th harmonic of string 2 at $300$ Hz. Below this, no common harmonic exists.

Final answer: difference in fundamentals $= 10$ Hz; lowest match at the 6th harmonic of string 1 / 5th harmonic of string 2 = $300$ Hz.

Why This Works

The frequencies of a fixed string form a harmonic series: $f, 2f, 3f, \ldots$. Two strings have a common frequency if one harmonic series intersects the other, which happens when the ratio of their fundamentals is a rational number $p/q$ — at the LCM of the appropriate harmonics.

Here the ratio is $50/60 = 5/6$, so smallest matching harmonics are $n_1 = 6, n_2 = 5$.

Alternative Method

Compute the LCM of $50$ and $60$ directly: LCM$(50, 60) = 300$. Then $300/50 = 6$ and $300/60 = 5$ give the harmonic numbers.