Two pulleys of negligible mass are connected to two blocks of mass $$m1$$ and $$m2$$ by two strings of the same material and the same $$cross-sectional$$ area as shown in Fig. The speed of a transverse wave in string AB is $$v1$$ and that in string CD is $$v2$$. The ratio $$v1v2is$$

Question

Two pulleys of negligible mass are connected to two blocks of mass $$m1$$ and $$m2$$ by two strings of the same material and the same $$cross-sectional$$ area as shown in Fig. The speed of a transverse wave in string AB is $$v1$$ and that in string CD is $$v2$$. The ratio $$v1v2is$$

Infinity Learn · Accepted Answer

Speed of transverse wave in a stretched string is $$v=T$$μ where T is the tension in the string and μ is the mass per unit length of the string. Since strings AB and CD have the same density and the same $$cross-sectional$$ area, they have the same μ. Hence v∝TIf the tension in CD is T, the tension in string AB is $$T2$$.∴ $$v1v2=TABTCD=T/2T=12So$$ the correct choice is (d).

Two pulleys of negligible mass are connected to two blocks of mass m₁ and m₂ by two strings of the same material and the same cross-sectional area as shown in Fig. The speed of a transverse wave in string AB is v₁ and that in string CD is v₂. The ratio $\frac{v_{1}}{v_{2}}$ is

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Two pulleys of negligible mass are connected to two blocks of mass m1 and m2 by two strings of the same material and the same cross-sectional area as shown in Fig. The speed of a transverse wave in string AB is v1 and that in string CD is v2. The ratio v1v2is

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Two pulleys of negligible mass are connected to two blocks of mass m₁ and m₂ by two strings of the same material and the same cross-sectional area as shown in Fig. The speed of a transverse wave in string AB is v₁ and that in string CD is v₂. The ratio $\frac{v_{1}}{v_{2}}$ is