JAMB Physics 2020 Objective — Question 25
Question 25 of 40 from the Joint Admissions and Matriculation Board (JAMB) Physics 2020 Objective paper, with the correct answer and a full explanation.
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25. A vibrating string has a tension of 40N and produces a note of 200Hz when plucked in the middle. If the length of the string is unaltered and the tension is increased to 160N, the frequency becomes
- A. A. 200Hz
- B. B. 400HzCorrect
- C. C. 800Hz
- D. D. 50Hz
Explanation
Frequency is proportional to the square root of tension (length unaltered). f2 = f1 x sqrt(T2/T1) = 200 x sqrt(160/40) = 200 x 2 = 400Hz.
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