Question

A wire of $9.8\times {10}^{-3}$kg per meter mass passes over a frictionless pulley fixed on the top of an inclined frictionless plane which makes an angle of ${30}^0$ 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 downwards. The whole system is in equilibrium. Now a transverse wave propagates along the wire with a velocity of $100m/sec$. Find the ratio of masses $M_{1}to{M}_2$.

Answer 1

when the system is in equilibrium

$M_{1}gsin\theta =T{M}_{2}g=T$

Now,the speed of a transverse wave in a wire of mass m per unit

length and stretched with a tension $T$ is given by

$V=\sqrt{\frac{T}{m}}\therefore T=V^{2}m$

given

$m=9.8\times {10}^{-3}Kg{m}^{-1}V=100m{s}^{-1}\therefore T={\left(100\right)}^2\times 9.8\times {10}^{-3}=98N\therefore M_1=\frac{T}{gsin\theta }=\frac{T}{gsin{30}^0}=20Kg$

and$M_2=\frac{T}{g}=10Kg\therefore \frac{M_1}{M_2}=\frac{20}{10}=\frac{2}{1}$.