One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
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Let cross section area of each arm is A
When a small pressure difference maintained between two column , then liquid falls through height y in one arm . now ( see the figure ) difference in liquid column in two arms = y + y = 2y
Let density of mercury = d
now,
Restoring force = - weight of Hg column in excess of one arm
= - ( volume × density ) × g
= - ( A × 2y × d ) g [mass = volume × density ]
= - 2Aydg
= - (2Adg) y
Compare with F = - mw²x [ restoring force]
mw² = 2Adg
w = √(2Adg/m)
2π/T = √(2Adg/m)
T = 2π√(m/2Adg)
Here m is mass of mercury column of length l
So, m = Ald
Hence, T = 2π√(Ald/2Adg)
T = 2π√(l/2g)
When a small pressure difference maintained between two column , then liquid falls through height y in one arm . now ( see the figure ) difference in liquid column in two arms = y + y = 2y
Let density of mercury = d
now,
Restoring force = - weight of Hg column in excess of one arm
= - ( volume × density ) × g
= - ( A × 2y × d ) g [mass = volume × density ]
= - 2Aydg
= - (2Adg) y
Compare with F = - mw²x [ restoring force]
mw² = 2Adg
w = √(2Adg/m)
2π/T = √(2Adg/m)
T = 2π√(m/2Adg)
Here m is mass of mercury column of length l
So, m = Ald
Hence, T = 2π√(Ald/2Adg)
T = 2π√(l/2g)
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Explanation:
One end of a U- tube containing mercury is connected to a suction pump and the other end is
connected to atmosphere. A small pressure difference is mentioned between the two columns. Show
that when the suction pump is removed, the liquid in the U-tube executes S.H.M?
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