A mass of 2.9 g is hanging underneath a spring which has an unknown spring constant. The period of oscillation is 5.5 seconds. Find the spring constant in N/m. Give your answer to 2 s.f.

T = 2 pi sqrt (m/k)

5.5 = 2 pi sqrt (.0029/k)
.87535 = sqrt (.0029/k)
.766= .0029/k

k = .0038 N/m

To find the spring constant, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is:

F = -kx

Where:
F is the force exerted by the spring (in Newtons),
k is the spring constant (in N/m),
x is the displacement from the equilibrium position (in meters).

In the case of simple harmonic motion (oscillation) of a mass hanging from a spring, the force exerted by the spring is also equal to the gravitational force acting on the mass:

F = mg

Where:
m is the mass (in kilograms),
g is the acceleration due to gravity (approximately 9.8 m/s²).

We can equate these two forces to find the relationship between the spring constant, the period of oscillation, and the mass:

mg = kx

To find the spring constant, we need to rearrange this equation. First, let's convert the mass from grams to kilograms:

m = 2.9 g = 2.9 × 10⁻³ kg

Now, we need to find the displacement, x. In simple harmonic motion, the displacement is related to the period of oscillation, T, by the equation:

T = 2π√(m/k)

Rearranging this equation, we can solve for the displacement:

x = (2πT)²(m/k)
x = (2πT)²m / k

Substituting this value of x into the equation mg = kx, we can solve for the spring constant:

mg = k(2πT)²m / k
g = (2πT)² / k
k = (2πT)² / g

Plugging in the given values: m = 2.9 × 10⁻³ kg, T = 5.5 s, and g = 9.8 m/s², we can calculate the spring constant, k:

k = (2π × 5.5)² / 9.8
k ≈ 197.17 N/m

Therefore, the spring constant is approximately 197.17 N/m (to 2 significant figures).