An object of specific gravity rho is hung from a thin steel wire. The fundamental frequency for transverse standing waves in wire is 300 Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency in Hz is
Answers
Answer:
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Explanation:
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Hence the new fundamental frequency is f' = 300 (2 p − 1 / 2p)
Explanation:
Fundamental frequency, f = 1 / 2L √ T / M = 1 / 2L √ pVg / M
Where, T→ Tension in air
V→ Volume
p→ Specific gravity
g→ Gravitational acceleration
M→ Linear density
Immersed in water to half of the volume
New Tension, T′ =T−upthrust force
⇒T′ = T− V / 2 ρg
⇒T′ = Vpg − V/2 g [ρ water =1 g/cc]
⇒T′ = Vg(p − 1 / 2 )
New fundamental frequency, f ' = 1 / 2L √ T' / M
T ′ = 1 / 2L √ Vg (p− 1 / 2 ) / M
f' / f = 1 / 2L √ Vg (p − 1 / 2 ) / M ÷ 1 / 2 L √ Vpg / M
f' / 300 = (p − 1 / 2 / p)
f' = 300 (2 p − 1 / 2p)
Hence the new fundamental frequency is f' = 300 (2 p − 1 / 2p)