IIT JEE , AIEEE Forum - Physics , Mechanics

next_gates

please explain me in detail...

a massless rod AB of length L is hung from two identical wires of equal length.A block of mass 'm' is attached at point O on the rod. find the value of AO so that a tuning fork excites the wire on the left in its fundamental tone and the wire on the right in its second harmonic.

ans. L/5

prateek2292

hey buddy,look.
since the tuning fork which vibrates the wires is same.
therefore,we can equate the two frequencies.
Let the length of two wires be'l'.
Let the tensions in the wire on left be T and on the right be t.
therefore,
after solving u will get T/t=4.
Now,
since the rod is in at rest,therefore we can equate the upward forces to the downward forces.
T+t=mg
4t+t=mg
5t=mg
t=mg/5
T=4mg/5.
Now, write the torque equation taking A as the centre.
Let the block of mass m be hanged at a distance x from A.
mgx-tL=0 (because it is equilibrium)
mgx=tL
mgx=mgL/5
Therefore,

x=L/5.

RATE ME IF U ARE SATISFIED BY MY ANS............

next_gates

pls tell me the method of solving T/t=4....thats wat i m not getting,,,

prateek2292

hey look
1/2l

T/m=2/2l

t/m

cancelling 2l on both sides,and squaring

T/m=4t/m

cancelling m,

T=4t

T/t=4

where m is the mass per lenght which is same for both the wires.

spy

i agree with u prateek

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