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 in Hz is
Answers
Answer:
Hey!
Case I : In air
Fundamental frequency, f2L1MT
= 2L1MpVg
Where, T→ Tension in air
V→ Volume
p→ Specific gravity
g→ Gravitational acceleration
M→ Linear density
Case II: Immersed in water to half of the volume
New Tension, T
′
=T−upthrust force
⇒T
′ =T− 2V ρg
Hope it will be helpful ✌️
Answer:
Case 1 (air)
Fundamental frequency:
$$n=\cfrac { 1 }{ 2L } \sqrt { \cfrac { T }{ m } } \\ \quad =\cfrac { 1 }{ 2L } \sqrt { \cfrac { V\rho }{ m } } $$
where
ρ= specific gravity
V= volume
m= linear density
Case II (immersed in water to half of the volume)
New tension:
$${ T }^{ ' }=T-upthrust\quad force\\ \quad =T-\cfrac { V }{ 2 } \rho g\\ \quad =Vg\left( \rho -\cfrac { 1 }{ 2 } \right) \quad \{ as\quad \rho (water)=1\quad g/c.c\} $$
∴ new fundamental frequency
$${ n }^{ ' }=\cfrac { 1 }{ 2L } \sqrt { \cfrac { { T }^{ ' } }{ m } } \\ \quad =\cfrac { 1 }{ 2L } \sqrt { \cfrac { Vg\left( \rho -\cfrac { 1 }{ 2 } \right) }{ m } } \\ \therefore \cfrac { { n }^{ ' } }{ n } =\cfrac { 1 }{ 2L } \sqrt { \cfrac { Vg\left( \rho -\cfrac { 1 }{ 2 } \right) }{ m } } \div \cfrac { 1 }{ 2L } \sqrt { \cfrac { V\rho g }{ m } } \\ or,\cfrac { { n }^{ ' } }{ 300 } =\sqrt { \cfrac { \rho -\cfrac { 1 }{ 2 } }{ \rho } } \\ \therefore { n }^{ ' }=300{ \left( \cfrac { 2\rho -1 }{ 2\rho } \right) }$$
Explanation:
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