IIT JEE , AIEEE Forum - Physics , Mechanics

kasirajan.1990

A block of mass M is placed on a smooth table . Its two sides are attached to fixed walls by means of collinear horizontal springs of spring constants k1 n k2 (k1>k2)
as shown in fig. The block is made to oscillate horizontally along the line of the two springs.wht is the frequency of its oscillations...?

Greatdreams

Is the answer 2

[M/(k1 + k2)]

I may be wrong

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editing a bit I calculated the time period
which is correct I think...didnt notice that its frequency

kasirajan.1990

nopes...the ans is 1/2pi rt(k1+k2)/m)

rhd92781

in these kind of arrangements K(eff) = K1 + k2
so f=1/2(pi) rt((k1+k2)/M)

Also, if block M is displaced by x leftwards, net force acting on M is
F=-(k1x + k2x)
both forces will try to bring back M to equilibrium
a=-(k1+k2)/M x
w^2 = (k1+k2)/M
f = 1/2(pi) rt((k1+k2)/M)

Greatdreams

Here is the procedure

Let the mass be disp by x

If suppose the spring with spring const be stretched by

the dist x then the other spring is compressed by a dist x

So the force exerted by 1st spring is - k1x while the

other exerts a force -k2x

So M d2x/dt2 = -(k1 + k2)x

So d2x/dt2 + (k1+k2)x/M = 0

This represents a SHM of freq = 1/2

(k1+k2)/M

kasirajan.1990

thnku so much palz...i needed the soln very urgently..thnx fr postin it so soon..!

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