(a) Example 1: Measure the unstretched length of the spring. Hang it with the object at rest and measure the stretched length. Call the difference in these lengths ∆x. Equating the weight of the object and the force exerted by the extended spring gives mg = k∆x from which k can be determined.
Example 2: Set the hanging mass into oscillation. Determine the period T by timing n oscillations and dividing that time by n. The equation T = 2π√ m / k can then be used to find k.
(b) The spring is stretched less when the object is at rest in the fluid. The fluid exerts an upward buoyant force on the object. Since the net force on the object is still zero, the spring does not need to exert as much force as before and thus stretches less.
c&d)
(1) Measure the length of the spring when the object is immersed in the liquid, and subtract the unstretched length to determine the amount the spring is stretched. This will allow calculation of the force exerted by the spring on the object.
(2) The volume of fluid displaced is equal to the volume of the object, which can be determined from the given mass and density of the object.
(3) The buoyant force on the object is equal to the difference of the object’s weight and the force exerted by the spring.
(4) The buoyant force also equals the weight of the displaced fluid, which equals the product of the fluid density, displaced volume, and g.
Symbol |
Quantity |
ρ |
fluid density |
V |
object volume = displaced water volume |
g |
acceleration of gravity |
m |
mass of object |
x |
spring stretch in air |
xw |
spring stretch in water |
First solving for k in air. mg = kx
Then in the fluid. Fsp = kxw
Fnet = 0
Fb = mg – Fsp
ρVg = mg –kxw
ρVg = mg – (mg/x) xw solve for ρ.