InterviewSolution
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InterviewSolution
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Length of string og musical instrument is varied from `L_(o)` to `2L_(o)` in 4 different cases. Wire is made of different materials of mass per unit length `mu,2mu,3mu,4mu` respectively. For first case (string -1) length is `L_(o)`, Tension is `T_(o)` then fundamental frequency is `f_(o),` for second case length of the string is `(3L_(o))/2` (`3^(rd)`Harmonic), for thid case length of the string is `(5L_(o))/4` (`5^(th)` Harmonic) and for the fouth case length of the string is `(7L_(o))/4` (`14^(th)` harmonic). if frequency of all is same then tension in string in terms of `T_(o)` will be: `{:((A),"String-1",(P),T_(0)),((B),"String-2",(Q),(T_(0))/(sqrt(2))),((C),"String-3",(R),(T_(o))/2),((D),"String-4",(S),(T_(o))/16),(,,(T),(3T_(o))/16):}` |
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Answer» Correct Answer - `(A) to P,(B) to R,(C) to T,(D) to S` Case -1 `" " L=L_(o), T=T_(o),f=f_(o)` `f_(1)=1/(2L_(o))sqrt((T_(o))/(mu))` Case-2: `L=(3L_(o))/2` `f_(2)=3/(2xx(3L_(o))/2)sqrt((T_(2))/(2mu))=f_(o)rArr f_(o)=1/(sqrt(2)L_(o))sqrt((T_(2))/(mu))rArr T_(2)=(T_(o))/2` Case-3: `L=(5L_(o))/4` `f_(3)=5/(2xx(5L_(o))/4)sqrt((T_(3))/(3mu))=f_(o) rArr f_(o)=2/(sqrt(3)L_(o))sqrt((T_(3))/(mu))rArr T_(3)=(3T_(o))/16` Case -4 : `L=(7L_(o))/4rArr f_(4)=14/(2xx(7L_(o))/4)sqrt((T_(4))/(4mu))=f_(o)rArr f_(o)=2/(L_(o))sqrt((T_(4))/(mu))rArr T_(4)=(T_(o))/16` |
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