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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Oscillations

A light string with a mass per unit length of 8.00 g/m has its ends tied to two walls separated by distance equal to three - fourth the length of the string .An object of mass m is suspended from the center of the string , putting a tension in the string then transverse wave speed in the string as a function of hanging mass

$(a)\;\sqrt{mg} \;m/sec\qquad(b)\;\large\frac{1}{4} \sqrt{mg \times 10^{3} } m/s\qquad(c)\;\large\frac{1}{8} \sqrt{5mg \times 10^{3} } m/s\qquad(d)\;None$

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Answer : (c) $\;\large\frac{1}{8} \sqrt{5mg \times 10^{3} } m/s\quad$Explanation : By F.B.D image $2 \times \large\frac{L}{2} cos \theta=\large\frac{3L}{4}$$cos \theta =\large\frac{3}{4}$$2 T sin \theta=mg$$2T \times \large\frac{4}{5}=mg$$T=\large\frac{5mg}{8}$$v=\sqrt{\large\frac{T}{4}}$$v=\sqrt{\large\frac{\large\frac{5 mg}{8}}{8 \times 10^{-3}}}$$v=\large\frac{1}{8}\;\sqrt{5mg \times 10^{3}}\;m/s\;$
answered Mar 10, 2014 by yamini.v
edited Mar 11, 2014 by yamini.v

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