A-wire-of-mass-9-8-10-3-kg-per-meter-passes-over-a-frictionless-pulley-fixed-on-the-top-of-an-inclined-frictionless-plane-which-makes-an-angle-of-30-with-the-horizontal-Masses-M-1-and-M-2-ar

Question Number 21870 by Tinkutara last updated on 05/Oct/17

A wire of mass 9.8 \times 10^{- 3} kg per meter

passes over a frictionless pulley fixed

on the top of an inclined frictionless

plane which makes an angle of 30 ° 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

downward . The whole system is in

equilibrium . Now a transverse wave

propagates along the wire with a

velocity of 100 m / s . Find M_{1} and M_{2}

(g = 9.8 m / s^{2}) .

Answered by ajfour last updated on 05/Oct/17

M_{1} g \sin 30 ° = M_{2} g

\Rightarrow M_{1} = 2 M_{2}

T e n s i o n i n w i r e b e T .

T = M_{2} g

v e l o c i t y o f t r a n s v e r s e w a v e

v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{M_{2} g}{μ}} = \sqrt{\frac{M_{2} (9.8 m / s^{2})}{9.8 \times 10^{- 3} k g / m}}

\Rightarrow {(100 m / s)}^{2} = M_{2} \times 10^{6} m^{2} / k g . s^{2}

o r M_{2} = 10 g

M_{1} = 2 M_{2} = 20 g .

Commented by Tinkutara last updated on 06/Oct/17

Thank you very much Sir!