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An object of specific gravity p is hung from a steel wire. The fundamentals frequency of the transverse standing wave is 300 hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency will be ?

An object of specific gravity p is hung from a steel wire. The fundamentals frequency of the transverse standing wave is 300 hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency will be ?

Grade:11

2 Answers

Vignesh
17 Points
6 years ago
Dear Nainesh,
We know that,fund. freq., n=(1/2l)*sqrt(mg/mu)
n(w) – n in water; n(o) n in air
2l- constant, mu constant. hence,
n(w)/n(o) = sqrt (wt. in air / wt. in water)
               = sqrt (mg/ (mg – Fb )  ; Fb – buoyant force
               = sqrt {v*d(o)*g)/[v*d(b)*g – (v/2)*d(w)*g]} ; d(o) – densityof object in air, d(w) – densityin water
               = sqrt {[d(b)/d(w)]/[d(b)/d(w) – ½ (1)]}   (cancelling and dividing by d(w) in num. and denom.)
               = sqrt [ p / (p – ½ )]
               = sqrt [2p / (2p- 1)]
hence, n(w) = n(o) * sqrt [2p / (2p – 1)] 
                   
Sakshi Kota
13 Points
6 years ago
Case I: ⎛⎝In air⎞⎠Fundamental frequency:n=12LTm−−√  =12LVPgm−−−−√Here,T is tension in airV is volume,P is specific gravityg is acceleration due to gravitym is linear densityCase II: ⎛⎝immersed in water to half of the volume⎞⎠New tension:T`=T−upthrust force    =T−V2ρg ⎡⎣ Density of water = ρ = 1 g/cc ⎤⎦    =VPg−V2g     =Vg(P−12)New fundamental Frequency:n`=12LT`m−−√  =12LVg(P−12)m−−−−−−−√Divide n` with n:n`n=12LVg(P−12)m−−−−−−−√÷12LVPgm−−−−√ n`300=Vg(P−12)m×mVPg−−−−−−−−−−−−√ n`=300(P−12P)     =300 (2P−12P

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