JAMB - Physics (1990 - No. 27)
A sonometer wire of length 100cm under a tension of 10N, has a frequency of 250Hz. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. The new tension is
5.1N
7.1N
14.0N
19.6N
Explanation
\(\frac{F_1}{F_2}\) = \(\sqrt\frac{T_1}{T_2}\)
\(\frac{250}{350}\) = \(\sqrt\frac{10}{T_2}\)
T2 = 19.6N
\(\frac{250}{350}\) = \(\sqrt\frac{10}{T_2}\)
T2 = 19.6N
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