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Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball (ball 2) is fixed at O, and the other (ball 1) is free. The balls are charged identically as a result of which the spring length increases `eta=2` times. Determine the change in frequency. |
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Answer» When the balls are unchangeed `v_(0)=(1)/(2 pi)sqrt((k)/(m))` where k is the force constant of the spring and ma is the mass of the oscillating ball (ball 1). When the balls are charged, ball 1 will oscillate about the new equilibrium position. At the equilibrium position of ball 1. `(1)/(4 pi epsilon_(0))(q^(2))/((eta l)^(2))=k(etal-l)=k(eta-1)` or `l^(3)=(q^(2))/(4 pi epsilon_(0)eta^(2)(eta-1)k)` ...(i) When ball 1 is displaced by a small distance from the equilibrium position to the right. the unbalanced froce to the reight is given by `F=(1)/(4 pi epsilon_(0))(q^(2))/((eta l+x)^(2))-k(eta l+x-1)` From Newton law, we have `m(d^(2)x)/(dt^(2))=(1)/(4 pi epsilon_(0))(q^(2))/(eta^(2)l^(2))[1+(x)/(eta l)]^(2)-kl(eta-1)-kx` `=(1)/(4 pi epsilon_(0))(q^(2))/(eta^(2)l^(2))[1-(2x)/(eta l)]-kl(eta-1)-kx` `=-[(1)/(4 pi epsilon_(0))(2q^(2))/(eta^(3)l^(3))+k]x` From Eqs (i) and (ii), `m(d^(2)x)/(dt^(2))=-[(2(eta-1))/(eta)k+k]x=-(3 eta-2)/(eta)kx` or `(d^(2)x)/(dt^(2))=-(3 eta-2)/(eta)(k)/(m)` or `a =-[(3eta-2 k)/(meta)]x` Comparing with `a=-omega^(2)x`, (i) `omega^(2)=(3 eta-2)/(eta)(k)/(m)` or `v=(1)/(2pi)sqrt(((3 eta-2)/(eta))(k)/(m))` (ii) or `(v)/(v_(0))=sqrt((3 eta-2)/(eta))` (iii) Thus the frequency is increased `sqrt((3 eta-2)//eta)` times . Hence `eta=2` and so frequency increases `sqrt(2)` times . |
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