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Explain the horizontal oscillations of a spring. |
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Answer» Solution :`(i)` Let `'m'` be the mass of a block attached to a mass less spring `(ii) 'K'` be the stiffness CONSTANT or force constant or spring constant of the spring. `(III) 'x_(0)'` be the mean or equilibrium position of the block. `(iv) 'x'` be the small displacement of the block towards right, and the released. `(V)` The block will oscillate back and forth about `x_(0)` `(vi)` If `F` be the restoring force, then it is proportional to the `'x'`    `F prop x` `F=-kx` `m(d^(2)x)/(DT^(2))=-kx` `(d^(2)x)/(dt^(2))=-(k)/(m)x` `omega^(2)=(k)/(m)` `omega=sqrt((k)/(m))rads^(-1)` `f=(omega)/(2pi)=(1)/(2pi)hertz` `T=(1)/(f)=2pisqrt((m)/(k))` second |
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