Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App