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A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass when the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. The value of m in kg is...........

Answer» <html><body><p>`(16)/(9)`<br/>`(9)/(16)`<br/>`(3)/(4)`<br/>`(4)/(3)`</p>Solution :Period of oscillation of <a href="https://interviewquestions.tuteehub.com/tag/spring-11487" style="font-weight:bold;" target="_blank" title="Click to know more about SPRING">SPRING</a> <br/> `T= 2pi sqrt((m)/(<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>))` <br/> `therefore 3= 2pi sqrt((m)/(k))"""…….."(1)` <br/> Periodic time when 1 kg is <a href="https://interviewquestions.tuteehub.com/tag/added-367625" style="font-weight:bold;" target="_blank" title="Click to know more about ADDED">ADDED</a> to m and <br/> `5= 2pi sqrt((m+1)/(k))"""…….."(2)` <br/> Taking ratio of eqn. (2) and (1) <br/> `(5)/(3) = sqrt((m+1)/(m))` <br/> `therefore (<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>)/(9)= (m+1)/(m)""` [squaring] <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/25m-298112" style="font-weight:bold;" target="_blank" title="Click to know more about 25M">25M</a> = 9m+9` <br/> `therefore 16m = 9` <br/> `therefore m= (9)/(16)`.</body></html>


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