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Sagot :
Answer:
i. 2.5 kHz ii. 5 kHz
Explanation:
The frequency of the wave in the rod is given by f = nv/2L where n = harmonic, v = speed of sound in rod = 5000 m/s and L = length of rod = 1 m
i. f1
For f1, n = 1
f1 = 1 × v/2L
= v/2L
Substituting the values of the variables into the equation, we have
= 5000 m/s ÷ (2 × 1 m)
= 5000 m/s ÷ 2 m
= 2500 /s
= 2500 Hz
= 2.5 kHz
ii. f2
For f2, n = 2
f2 = 2 × v/2L
= v/L
Substituting the values of the variables into the equation, we have
= 5000 m/s ÷ 1 m
= 5000 /s
= 5000 Hz
= 5 kHz
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