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A smooth wedge of mass m and angle of inclination 60^(@)rests unattached between two springs of spring constant K and 4k on a smooth horizontal plane both springs in the unextended position the time period of small oscillation of the wedge (assuming that the springs are constrained to get compressed along their length) equals |
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Answer» `pi(1+1/2)SQRT(m/k)`  `F=4kx cos^(2)30^(@) Rightarrow VEC(a) =-((k3)/M^(4))xx4vec(x) Rightarrow infty^(2)=(3k)/M infty=sqrt((3k)/M)` `Rightarrow T=(2pi)/infty=(2pi)/sqrt(3k//M) Rightarrow T=(2pi)/sqrt(3k)sqrt(M) Rightarrowt_(1)=T_(1)/2=pi/sqrt(3k) sqrt(M)` `Rightarrow t_(2)=T_(2)/2=pisqrt(M/k)` Rightarrow time period `= t_(1)+ t_(2) = [(pisqrt(M))/sqrt(3k)+pi sqrt(M/k)]` `Rightarrow time period= t_(1)+t_(2)= pisqrt(m/k)[1+1/sqrt(3)]` ANS |
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