A) 5 m/s, 2.24 m/s
B) 10 m/s, 3.23 m/s
C) 7.5 m/s, 1.2 m/s
D) none of these
Correct Answer: A
Solution :
Mass per unit length \[m=\frac{M}{L}\] Tension at the point T= mass of \[x\] meter of rope \[\times \] \[g=mx\text{ }g\] where \[x\] is the distance of point from lower end Velocity of transverse wave along the string \[\upsilon =\sqrt{\frac{T}{m}}=\sqrt{x\frac{mg}{m}}=\sqrt{xg}\] Here: \[x=0.5\,m\] \[g=10\,m/{{s}^{2}}\] \[\upsilon =\sqrt{0.5\times 10}=\sqrt{5}=2.24\,m/s\] At upper end, the velocity is given by \[\upsilon =\sqrt{rg}=\sqrt{2.5\times 10}=5\,m/s\]
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