question_answer

A wire of mass \[\text{9}\text{.8}\times \text{l}{{0}^{-3}}\text{kg}\] per meter passes over a frictionless pulley fixed on the top of an inclined frictionless plane which makes an angle of \[\text{3}0{}^\circ \] with the horizontal. Masses \[{{M}_{1}}\] and \[{{M}_{2}}\] are tied at the two ends of the wire. The mass\[{{M}_{1}}\] rests on the plane and the mass \[{{M}_{2}}\] hangs freely vertically downward. The whole system is in equilibirum. Now a transverse wave1 propagates along the wire with a velocity of 100 m/s. Find \[{{M}_{1}}\] and \[{{M}_{2}}\] (\[\text{g}=\text{9}\text{.8 m}/{{\text{s}}^{\text{2}}}\]).

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),            \[{{M}_{1}}=2{{M}_{2}}=2\times 10=20kg\]