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 produ...

Assessment: JAMB UTME - Physics - 1990 Subject: Physics

Question 1 Report

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
Answer Details
The tension and frequency of a sonometer wire are directly proportional, meaning that an increase in tension results in an increase in frequency, and vice versa, as long as the length of the wire is kept constant. We can use the formula: f = (1/2L) * sqrt(T/μ) where: - f is the frequency of the wire - L is the length of the wire - T is the tension in the wire - μ is the mass per unit length of the wire If we assume that μ is constant and the length of the wire is 100cm, we can set up two equations for the two given frequencies: 250 = (1/2 * 100) * sqrt(10/T) 350 = (1/2 * 100) * sqrt(x/T) where x is the new tension we want to find. Simplifying the equations, we get: sqrt(10/T) = 5/2 sqrt(x/T) = 7/2 Squaring both sides of each equation, we get: 10/T = 25/4 x/T = 49/4 Multiplying both sides of each equation by T, we get: 10 = 25/4 * T x = 49/4 * T Solving for T in the first equation, we get: T = 10 * 4/25 = 1.6 N Substituting this value of T into the second equation, we get: x = 49/4 * 1.6 = 19.6 N Therefore, the new tension required to produce a frequency of 350 Hz is 19.6 N.

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