A string of length L = 1 m is fixed at both ends — JEE Physics
A string of length $L = 1$ m is fixed at both ends. Find the first three harmonics (natural frequencies) if the wave speed is $v = 200$ m/s.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 16d ago
▲ 34
For fixed-fixed string, $\lambda_n = \dfrac{2L}{n}$, so $f_n = \dfrac{nv}{2L}$:
$$f_1 = \frac{1 \times 200}{2} = 100 Hz$$
$$f_2 = 200 Hz, \quad f_3 = 300 Hz$$
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Discussion (3)
K
Can someone explain why we ignore the other root here?
SJ
Thanks a ton, I was stuck on this exact problem for an hour.
VA
Good follow-up questions — remember to always state your assumptions in the JEE subjective section.