Answer:
Here,
mass/length, \[\text{m}=\text{9}\text{.8}\times
\text{1}{{0}^{-3}}\text{kg }{{\text{m}}^{\text{-1}}}\]
\[\theta ={{30}^{o}}\], \[g=9.8m/{{s}^{2}}\]
\[\upsilon =100m{{s}^{-1}}\]; \[{{M}_{1}}=?\] \[{{M}_{2}}=?\]
The various
forces acting on the system are shown in Fig. 3(HT).9. As the system of two
masses is in equilibrium
\[\therefore \] \[T={{M}_{1}}g\,\sin \theta
={{M}_{1}}g\sin {{30}^{o}}\] ??.. (i)
\[R={{M}_{1}}g\,co\operatorname{s}\theta
={{M}_{1}}g\cos {{30}^{o}}\] ??..(ii)
\[T={{M}_{2}}g\] ??..(iii)
From (i) and (iii),
\[{{M}_{1}}g\,\sin
{{30}^{o}}={{M}_{2}}g\]
\[\frac{{{M}_{1}}}{2}={{M}_{2}}\] or
\[{{M}_{1}}=2{{M}_{2}}\]
Now, the velocity v of
transverse wave is given by
\[\upsilon =\sqrt{\frac{T}{m}}\] or \[T={{\upsilon
}^{2}}\times m={{\left( 100 \right)}^{2}}\times 9.8\times {{10}^{-3}}=98N\]
From (iii), \[{{M}_{2}}=\frac{T}{g}=\frac{98}{9.8}=10kg\]
From (iv), \[{{M}_{1}}=2{{M}_{2}}=2\times
10=20kg\]
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