Grade 11Wave Motion

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 ?

Nainesh advani

8 Years agoGrade 11

2 Answers

Vignesh

8 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

8 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