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A load of mass m falls from a height h on to the scale pan hung from a spring as shown in the adjoining figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is

Answer» <html><body><p>`(mg)/(k)`<br/>`(mg)/(k)<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(((1+2hk)/(mg)))`<br/>`(mg)/(k)+(mg)/(k)sqrt(((1+2hk)/(mg)))`<br/>`(mg)/(k)sqrt(((1+2hk)/(mg)-(mg)/(k)))`</p>Solution :According to conservation principle <br/> `mgh=(1)/(2)kx_(0)^(2)-mgx_(0)` <br/> where, `x_(0)` is maximum elongation in spring (when <a href="https://interviewquestions.tuteehub.com/tag/particle-1147478" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICLE">PARTICLE</a> is in its <a href="https://interviewquestions.tuteehub.com/tag/lowest-1081124" style="font-weight:bold;" target="_blank" title="Click to know more about LOWEST">LOWEST</a> <a href="https://interviewquestions.tuteehub.com/tag/extreme-981946" style="font-weight:bold;" target="_blank" title="Click to know more about EXTREME">EXTREME</a> position). <br/> or `(1)/(2)kx_(0)^(2)-mgx_(0)-mgh=0` <br/> or `x_(0)^(2)-(2mg)/(k)x_(0)-(2mg)/(k)h=0` <br/> `therefore""x_(0)=((2mg)/(k)pmsqrt([((2mg)/(k))^(2)+4xx(2mg)/(k)h]))/(2)` <br/> `because` Amplitude=elongation in spring for lowest extreme position-elongation in spring for equilibrium position <br/> `=x_(0)-x_(1)=(mg)/(k)sqrt((1+(2hk)/(mg)))`</body></html>


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