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A-sonometer-wire-60cm-long-under-a-tension-of-120N-vibrates-in-unison-with-a-turning-fork-of-frequency-512Hz-The-wire-is-then-shortened-by-2cm-the-tension-remaining-the-same-If-the-fork-and-wire-are-s




Question Number 35878 by NECx last updated on 25/May/18
A sonometer wire 60cm long,under  a tension of 120N,vibrates in  unison with a turning fork of  frequency 512Hz.The wire is then  shortened by 2cm,the tension  remaining the same.If the fork  and wire are sounded together,how  many beats per second are heard?  What decrease in tension will be  required to restore the frequency  of vibration of the wire to the  original value?
$${A}\:{sonometer}\:{wire}\:\mathrm{60}{cm}\:{long},{under} \\ $$$${a}\:{tension}\:{of}\:\mathrm{120}{N},{vibrates}\:{in} \\ $$$${unison}\:{with}\:{a}\:{turning}\:{fork}\:{of} \\ $$$${frequency}\:\mathrm{512}{Hz}.{The}\:{wire}\:{is}\:{then} \\ $$$${shortened}\:{by}\:\mathrm{2}{cm},{the}\:{tension} \\ $$$${remaining}\:{the}\:{same}.{If}\:{the}\:{fork} \\ $$$${and}\:{wire}\:{are}\:{sounded}\:{together},{how} \\ $$$${many}\:{beats}\:{per}\:{second}\:{are}\:{heard}? \\ $$$${What}\:{decrease}\:{in}\:{tension}\:{will}\:{be} \\ $$$${required}\:{to}\:{restore}\:{the}\:{frequency} \\ $$$${of}\:{vibration}\:{of}\:{the}\:{wire}\:{to}\:{the} \\ $$$${original}\:{value}? \\ $$
Commented by NECx last updated on 28/May/18
somebody please help
$${somebody}\:{please}\:{help} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18
f_1 =(1/(2l_1 ))(√(T/m))  f_1 =512   l_1 =0.6meter  T=120N  f_2 =(1/(2l_2 ))(√(T/m))  (f_2 /f_1 )=(l_1 /l_2 )  f_2 =f_1 ×(l_1 /l_2 )  f_2 =512×((0.6)/(0.58))  f_2 =((512×60)/(58))=530  beats=530−512=18  m=(T/(4l_1 ^2 f_1 ^2 ))  512=(1/(2×0.58))×(√((T′)/m))  f_1 =(1/(2×l_2 ))×(√((T′×4l_1 ^2 f_1 ^2 )/T))  1=(l_1 /l_2 )×(√((T′)/T))  T′=T×(l_2 ^2 /l_1 ^2 )  T′=120×((0.58×0.58)/(0.60×0.60))=112  △T=120−112=8N
$${f}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{l}_{\mathrm{1}} }\sqrt{\frac{{T}}{{m}}} \\ $$$${f}_{\mathrm{1}} =\mathrm{512}\:\:\:{l}_{\mathrm{1}} =\mathrm{0}.\mathrm{6}{meter}\:\:{T}=\mathrm{120}{N} \\ $$$${f}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}{l}_{\mathrm{2}} }\sqrt{\frac{{T}}{{m}}} \\ $$$$\frac{{f}_{\mathrm{2}} }{{f}_{\mathrm{1}} }=\frac{{l}_{\mathrm{1}} }{{l}_{\mathrm{2}} } \\ $$$${f}_{\mathrm{2}} ={f}_{\mathrm{1}} ×\frac{{l}_{\mathrm{1}} }{{l}_{\mathrm{2}} } \\ $$$${f}_{\mathrm{2}} =\mathrm{512}×\frac{\mathrm{0}.\mathrm{6}}{\mathrm{0}.\mathrm{58}} \\ $$$${f}_{\mathrm{2}} =\frac{\mathrm{512}×\mathrm{60}}{\mathrm{58}}=\mathrm{530} \\ $$$${beats}=\mathrm{530}−\mathrm{512}=\mathrm{18} \\ $$$${m}=\frac{{T}}{\mathrm{4}{l}_{\mathrm{1}} ^{\mathrm{2}} {f}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\mathrm{512}=\frac{\mathrm{1}}{\mathrm{2}×\mathrm{0}.\mathrm{58}}×\sqrt{\frac{{T}'}{{m}}} \\ $$$${f}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}×{l}_{\mathrm{2}} }×\sqrt{\frac{{T}'×\mathrm{4}{l}_{\mathrm{1}} ^{\mathrm{2}} {f}_{\mathrm{1}} ^{\mathrm{2}} }{{T}}} \\ $$$$\mathrm{1}=\frac{{l}_{\mathrm{1}} }{{l}_{\mathrm{2}} }×\sqrt{\frac{{T}'}{{T}}} \\ $$$${T}'={T}×\frac{{l}_{\mathrm{2}} ^{\mathrm{2}} }{{l}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${T}'=\mathrm{120}×\frac{\mathrm{0}.\mathrm{58}×\mathrm{0}.\mathrm{58}}{\mathrm{0}.\mathrm{60}×\mathrm{0}.\mathrm{60}}=\mathrm{112} \\ $$$$\bigtriangleup{T}=\mathrm{120}−\mathrm{112}=\mathrm{8}{N} \\ $$
Commented by NECx last updated on 29/May/18
wow..... Thanks mr Tanmay.
$${wow}…..\:{Thanks}\:{mr}\:{Tanmay}. \\ $$$$ \\ $$

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