Question

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 wave are $n_1,n_{2}and{n}_3$. Then

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

Answer 1

The correct option is D $n_2=\frac{f_1+f_2}{2}=75Hz$

We know that: $f\propto \frac{1}{\lambda }$

${\lambda }_1=2l$

${\lambda }_2=l$

${\lambda }_3=\frac{2l}{3}$

Clearly, $\frac{2}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$

$\Rightarrow 2n_3=n_1+n_2$

$\Rightarrow n_3=\frac{n_1+n_2}{2}$