A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg/m³ is supported by a vertical spring and is half dipped in water.
a) Find the elongation of the spring in the equilibrium condition. (b) If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant = 500 N/m.
Answers
(a) . A block is supported in water by the Spring of spring constant(k) = 500 N/m.
Now, Upthrust and Spring Force acts in Upwards direction and weight acts in downwards direction.
∴ Upthrust by Fluid + Spring Force = Weight
∴ Vρg + kx = mg
[V is the volume of solid immersed and ρ is the density of the fluid.]
Now, Vρg = πr²h/2 × 1000 × 9.8
= 22/7 × (5/100)²× (20/200) × 1000 × 9.8
= 22/7 × 25/10000 × 1/10 × 1000 × 9.8
= 22/7 × 25/100 × 9.8
= 7.7 N
Now, mg = density of cylindrical object × volume of cylindrical object × g
= 8000 × 22/7 × (10/200)² × (20/100) × 9.8
= 8000 × 22/7 × 1/400 × 1/5 × 9.8
= 20/5 × 22/7 × 9.8
= 4 × 9.8 × 22/7
= 123.2 N.
∴ kx = 123.2 - 7.7
500x = 115.5
∴ x = 0.231 m.
∴ x = 23.1 cm.
Hence, the Elongation in the spring is 23.1 cm.
______________________________
(b). For finding the time period, I uses three steps.
Step 1. ⇒ Find the equilibrium Position.
kx₀ + Vρg = mg
Step 2. ⇒ Displace the Object slight x and then find new F net. Then Mathematically arrive at F = kx.
kx = mg - V'ρg
kx = mg - A(x)ρg
mg = kx + Axρg
F = x(k + Aρg)
Now, Compare it with F = kx,
k' = k + Aρg
Therefore,
T = 2π/w
T = 2π√(m/k')
T = 2π√[m/(k + Aρg)]
∴ T = 2 × 22/7 × √[m/(k + Aρg)]
Now, Let us First find the Value of the m/(k + Aρg).
∴ m = density × volume
∴ m = 8000 × πr²h
∴ m = 8000 × 22/7 × (5/100)² × (20/100)
∴ m = 12.6 kg.
k + Aρg = 500 + πr² × g
= 500 + 22/7 × (5/100)² × 9.8
= 500 + 0.077
= 500.077
∴ T = 2 × 22/7 × √(12.6/500.077)
∴ T = 0.99 seconds.
Hope it helps.
The volume of the object =π*5²*20 cm³ =500π cm³
The mass of the object =500π*8000/10⁶ kg
=4π kg
Since the object is half dipped, the volume of the water displaced =500π/2 =250π cm³
The mass of the water displaced
=250π*1000/10⁶ kg
=0.25π kg
The force of buoyancy =0.25πg N
Hence the apparent weight of the object
=(4π-0.25π)g N
=3.75πg N
The spring constant k =500 N/m
(a) Hence the elongation of the spring =3.75πg/500 m
=3.75*3.14*10/500 m {Taking g = 10 m/s²}
=0.235 m
=23.5 cm
(b) Let the object be dipped to a distance X m. Assuming the axis of the cylindrical object vertical, the extra volume displaced =π*5²*X/10000 m³
The buoyancy force =(π*5²*X/10000 m³)*1000g N
=2.5πgX N
The spring force =kX N
Hence the total upward force =2.5πgX+500X N
Hence the vertical acceleration at the moment =Force/mass
=(2.5πgX+500X)/4π m/s²
=(2.5πg+500)X/4π m/s²
But also this acceleration =k²X m/s²
Equating, k²X = (2.5πg+500)X/4π
=46.04X
→k=√(46.04) =6.785 s⁻¹
So, the time period =2π/k
=2π/6.785 s =0.93 s