$A$ string of length $1\,m$ and linear mass density $0.01\,kg/m$ is stretched to a tension of $100\,N$. When both ends of the string are fixed,the three lowest frequencies for standing waves are $f_1, f_2$,and $f_3$. When only one end of the string is fixed,the three lowest frequencies for standing waves are $n_1, n_2$,and $n_3$. Then:

  • A
    $n_3 = 5n_1 = f_3 = 125\,Hz$
  • B
    $f_3 = 5f_1 = n_2 = 125\,Hz$
  • C
    $f_3 = n_2 = 3f_1 = 150\,Hz$
  • D
    $n_2 = \frac{f_1 + f_2}{2} = 75\,Hz$

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