A wooden block of mass 0.5 kg and density 800 kg/m³ is fastened to the free end of a vertical spring of spring constant 50 N/m fixed at the bottom. If the entire system is completely immersed in water, find (a) the elongation (or compression) of the spring in equilibrium and (b) the time period of vertical oscillations of the block when it is slightly depressed and released.
Answers
(a). When the system is completely Immersed in a liquid, then Upthrust and the spring force acts vertically upwards, but the weight of the object acts downwards,
Thus, In the State of the Equilibrium, we can say that,
Spring force + Upthrust = Weight
kx + Vρg = mg
∴ kx = mg - Vρg
∴ kx = 0.5 × 10 - 0.5/800 × 1000 × 10
∴ kx = 5 - 50/8
kx = -1.25 [Minus sign means upward direction.]
∴ x = 0.025 m or 2.5 cm.
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(b). Let the block be depressed by 'a' cm.
The force on the block by the spring =50a N
Hence the acceleration = 50a/0.5
= 100a m/s²
Now, ⍵²a = 100
∴ ⍵ = 10 s⁻¹
Using the formula,
T = 2π/⍵
∴ T =2π/10
∴ T = π/5 seconds or 0.63 seconds.
Hence, the time period of the S.H.M. is π/5 or 0.63 seconds.
Hope it helps.
Answer:
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