WAEC - Physics (1998 - No. 34)
A sonometer wire under a tension of N10N, produces a frequency of 250Hz when plucked. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. Calculate the new tension
39.2N
19.6N
14.2N
7.4N
5.1N
Explanation
f \(\alpha\) \(\sqrt{T}\)
\(\frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}}\)
\(\frac{250}{350} = \sqrt{\frac{10}{T_2}}\)
\(\frac{5}{7} = \sqrt{\frac{10}{T_2}}\)
\(\frac{25}{49} = \frac{10}{T_2}\)
T2 = \(\frac{49 \times 10}{25} = 19.6\)
\(\frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}}\)
\(\frac{250}{350} = \sqrt{\frac{10}{T_2}}\)
\(\frac{5}{7} = \sqrt{\frac{10}{T_2}}\)
\(\frac{25}{49} = \frac{10}{T_2}\)
T2 = \(\frac{49 \times 10}{25} = 19.6\)
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