Question
Two strings $$A$$ and $$B$$ have lengths $${l_A}$$ and $${l_B}$$ and carry masses $${M_A}$$ and $${M_B}$$ at their lower ends, the upper ends being supported by rigid supports. If $${n_A}$$ and $${n_B}$$ are the frequencies of their vibrations and $${n_A} = 2\,{n_B},$$ then
A.
$${l_A} = 4{l_B},$$ regardless of masses
B.
$${l_B} = 4{l_A},$$ regardless of masses
C.
$${M_A} = 2{M_B},{l_A} = 2{l_B}$$
D.
$${M_B} = 2{M_A},{l_B} = 2{l_A}$$
Answer :
$${l_B} = 4{l_A},$$ regardless of masses
Solution :
The frequency of vibrations of string is
$$\eqalign{
& n = \frac{1}{2}\sqrt {\frac{g}{l}} \,......\left( {\text{i}} \right) \cr
& {\text{Given,}}\,\,{n_A} = 2{n_B} \cr
& \therefore \frac{1}{2}\sqrt {\frac{g}{{{l_A}}}} = 2 \cdot \frac{1}{2}\sqrt {\frac{g}{{{l_B}}}} \cr
& {\text{or}}\,\,\frac{1}{{{l_A}}} = \frac{4}{{{l_B}}}\,\,or\,\,{l_B} = 4{l_A} \cr} $$
It is obvious from Eq. (i), the frequency of vibrations of strings does not depend on their masses.