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| 1. |
Describe the vertical oscillations of a spring. |
| Answer» <html><body><p></p>Solution :Vertical oscillations of a spring: Let us consider a massless spring with stiff <a href="https://interviewquestions.tuteehub.com/tag/ness-1113753" style="font-weight:bold;" target="_blank" title="Click to know more about NESS">NESS</a> constant or force constant k attached to a ceiling as shown in figure. Let the <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring <a href="https://interviewquestions.tuteehub.com/tag/elongates-969691" style="font-weight:bold;" target="_blank" title="Click to know more about ELONGATES">ELONGATES</a> by a length l. Let `F_(1)` be the restoring force due to stretching of spring. Due tomass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium, <br/> `F_(1)+mg=0""...(1)` <br/> But the spring elongates by small displacement l, therefore, <br/> `F_(1)proplimpliesF_(1)=-<a href="https://interviewquestions.tuteehub.com/tag/kl-528180" style="font-weight:bold;" target="_blank" title="Click to know more about KL">KL</a>""...(2)` <br/> Substituting equation (2) in equation (1), we get <br/> `-kl+mg=0""...(3)` <br/> `mg=kl" (or) "(m)/(k)=(l)/(g)""...(3)` <br/> Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is `y+1`) is <br/> `F_(2)prop(y+l)` <br/> `F_(2)=-k(y+l)=-ky-kl""...(4)` <br/> Since, the mass moves up and down with acceleration `(d^(2)y)/(dt^(2))`, by drawing the free body diagram for this case, we get <br/> `-ky-kl+mg=m(d^(2)y)/(dt^(2))""...(5)` <br/> The net force acting on the mass due to this stretching is <br/> `F=F_(2)+mg` <br/> `F=-ky-kl+mg""...(6)` <br/> The gravitational force opposes the restoring force. Substituting equation (3) in equation (6), we get <br/> `F=-ky-kl+kl=-ky` <br/> Applying Newton.s law, we get <br/> `m(d^(2)y)/(dt^(2))=-kyimplies (d^(2)y)/(dt^(2))=-(k)/(m)y"" ...(7)` <br/> The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as <br/> `T=2pisqrt((m)/(k))" second"...(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)` <br/> The time period can be rewritten using equation (3) <br/> `T=2pisqrt((m)/(k))=2pisqrt((l)/(g))""...(9)` <br/> The acceleration due to gravity g can be computed from the formula <br/> `g=4pi^(2)((1)/(T^(2)))ms^(-2)""...(10)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V02_C10_E01_037_S01.png" width="80%"/></body></html> | |