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A wire of density 'rho' and Yaung's modulus 'Y' is stretched between two rigid supports separated by a distance 'L' under tension 'T'. Derive an expression for its frequency if fundamental mode. Hence show that n=(1)/(2L)sqrt((Yl)/(rhoL)), where symbolshave their usual meanings. When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length of the pendulum. |
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Answer» Solution :Consider a wire stretched between TWO rigid supports ata distance L apart. Let T = the TENSION in the wire r = the radius of cross section of the wire. Y, `rho` = young's modulus and mass denisty of the material of the wire. M, m = the mass and linear density of the wire Then, `M=(pir^(2)L)rho" and "m=(M)/(L)=pir^(2)rho""...(i)` `because` The stress in the wire `=(T)/(pir^(2))` `:.""(T)/(m)=(T)/(pir^(2)rho)=("Stress")/(rho)""...(ii)` The fundamental frequency of vibration of the wire, `n=(1)/(2L)sqrt((T)/(m))` `=(1)/(2L)sqrt(("Stress")/(rho))""["Using"(ii)]...(iii)` If `DeltaL=l` is the elastic extension of the wire under tension T, Strain =l/L Since `""Y=("Stress")/("Strain")` `implies"""Stress"=Yxx"strain"=Y(l)/(L)""...(iv)` `:.""n=(1)/(2L)sqrt((Yl)/(rhoL))""...(v)` which is the required expression. Numerical : Given : Length decreasedby 20 CM and PERIOD changes by `10%`. Let, original length = l andnew time period will decreased by `10%` `T'=T-(10)/(100)xxT` `:.""T'=T-(T)/(10)=(9T)/(10)` `because""T=2pisqrt((l)/(g))""...(i)` `implies""(9T)/(10)=2pisqrt((l-20)/(g))""...(ii)` On dividing equation (i) by (ii) `implies""(T)/((9T)/(10))=(2pisqrt((l)/(g)))/(2pisqrt((l-20)/(g)))` `implies""(10)/(9)=sqrt((l)/(l-20))` `implies""((10)/(9))^(2)=(l)/(l-20)` `implies""(100)/(81)=(l)/(l-20)` `implies""100(l-20)=81l` `implies""100l-81l=2000` `implies""19l=2000` `implies""l=(2000)/(19)" cm"` `=(20)/(9)" m"` `:.""l=1.05m` |
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