Two masses MA and MB are hung from two strings of length lA and lB respectively. They are executing SHM with frequency relation fA=2 fB, then relation

Question

Two masses MA and MB are hung from two strings of length lA and lB respectively. They are executing SHM with frequency relation fA=2 fB, then relation (A) lA=(lB/4), does not depend on mass (B) lA=4 lB, does not depend on mass (C) lA=2 lB and MA=2 MB (D) lA=(lB/2) and MA=(MB/2).

Tardigrade Admin · Accepted Answer

fA=2 fB ⇒ (1/2 π) √(g/lA)=2 × (1/2 π) √(g/lB) or, (1/lA)=4 × (1/lB). or, lA=(lB/4), which does not depend on mass.

Q. Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing SHM with frequency relation $f_{A} = 2 f_{B}$ , then relation

$l_{A} = \frac{l _{B}}{4}$ , does not depend on mass

$l_{A} = 4 l_{B}$ , does not depend on mass

$l_{A} = 2 l_{B}$ and $M_{A} = 2 M_{B}$

$l_{A} = \frac{l _{B}}{2}$ and $M_{A} = \frac{M _{B}}{2}$ .

$f_{A} = 2 f_{B}$
$\Rightarrow \frac{1}{2 π} \frac{g}{l _{A}} = 2 \times \frac{1}{2 π} \frac{g}{l _{B}}$
or, $\frac{1}{l _{A}} = 4 \times \frac{1}{l _{B}}$ .
or, $l_{A} = \frac{l _{B}}{4}$ ,
which does not depend on mass.

Q. Two masses MA​ and MB​ are hung from two strings of length lA​ and lB​ respectively. They are executing SHM with frequency relation fA​=2fB​, then relation

lA​=4lB​​, does not depend on mass

lA​=4lB​, does not depend on mass

lA​=2lB​ and MA​=2MB​

lA​=2lB​​ and MA​=2MB​​.