A string is tied between the two poles separated by the distance 3 m to demonstrate the standing wave experiment. consider the length of the string is 100 m and mass is 0.15kg for which the second harmonic frequency is 35HZ. calculate the tension in the string?
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Concept:
- Standing wave
- A standing wave, sometimes called a stationary wave in physics, oscillates in time but has a fixed peak amplitude profile in space.
- fundamental frequency
- nth order harmonic
Given:
- Distance between poles L = 3 m
- Mass per unit length μ = 0.15/100 = 0.0015 kg/m
- Second harmonic frequency f2 = 35 Hz
Find:
- The tension in the string
Solution:
The nth order frequency is given by
fn = (n/2L) √T/μ
f2 = (2/2L)√T/μ
f2 = (1/L)√T/μ
f2 = 35, L = 3 m, μ = 0.0015 kg/m
35 = 1/3 √T/0.0015
35*3 = √T/0.0015
√T/0.0015 = 35*3
√T/0.0015 = 105
T/0.0015 = 105^2
T/0.0015 = 11025
T = 16.5 N
The tension in the string is 16.5 N.
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