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

$${A}\:{sonometer}\:{wire}\:\mathrm{60}{cm}\:{long},{under} \\ $$$${a}\:{tension}\:{of}\:\mathrm{120}{N},{vibrates}\:{in} \\ $$$${unison}\:{with}\:{a}\:{turning}\:{fork}\:{of} \\ $$$${frequency}\:\mathrm{512}{Hz}.{The}\:{wire}\:{is}\:{then} \\ $$$${shortened}\:{by}\:\mathrm{2}{cm},{the}\:{tension} \\ $$$${remaining}\:{the}\:{same}.{If}\:{the}\:{fork} \\ $$$${and}\:{wire}\:{are}\:{sounded}\:{together},{how} \\ $$$${many}\:{beats}\:{per}\:{second}\:{are}\:{heard}? \\ $$$${What}\:{decrease}\:{in}\:{tension}\:{will}\:{be} \\ $$$${required}\:{to}\:{restore}\:{the}\:{frequency} \\ $$$${of}\:{vibration}\:{of}\:{the}\:{wire}\:{to}\:{the} \\ $$$${original}\:{value}? \\ $$

Commented by NECx last updated on 28/May/18

$${somebody}\:{please}\:{help} \\ $$

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

$${wow}…..\:{Thanks}\:{mr}\:{Tanmay}. \\ $$$$ \\ $$

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