A sonometer wire, under a tension of 22.5 N. produces 4 beats per second with a tuning fork. When the tension in the wire is increased to 22.8 N, it produces 2 beats per second with the fork. Find the frequency of the fork.
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Explanation:
Given A sonometer wire, under a tension of 22.5 N. produces 4 beats per second with a tuning fork. When the tension in the wire is increased to 22.8 N, it produces 2 beats per second with the fork. Find the frequency of the fork.
- So frequency is proportional to square root of tension in the string
- So f α √T
- Therefore f1 / f2 = √T1 / T2
- Here T1 = 22.5 N and T2 = 22.8 N
- Therefore T1 < T2 and f1 < f2
- In the first case 4 beats / sec is produced
- So frequency of tuning fork is either f + 4 or f – 4
- In the second case 2 beats / sec is produced
- So frequency of tuning fork is either f + 2 or f – 2
- Now consider f – 4 and f + 2
- f1 / f2 = √T1 / T2
- f – 4 / f + 2 = √22.5 / 22.8
- = √225 / 228
- f – 4 / f + 2 = 15 / 15.09966
- = 0.993399
- f – 4 = (f + 2)0.993399
- f – 4 = 0.993399 f + 1.9867997028
- Or f = 5.9867997028 / 0.006601
- Or f = 906 hertz
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