Question

A block of mass m= 6.00 kg is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h= 5.00 cm from its equilibrium length. (Figure 1)The spring has an unknown spring constant k. Take the acceleration due to gravity to be g = 9.81 m/s2 .

Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting frequency f of the block's oscillations about its equilibrium position.
Express your answer in hertz.

Answer 1

To find the resulting frequency of the block's oscillations about its equilibrium position, first find the spring constant using Hooke's Law and then use the formula for the frequency of a mass-spring system.

Step-by-step explanation:

To find the resulting frequency of the block's oscillations about its equilibrium position, we first need to find the spring constant (k) of the ideal spring.

To do this, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.

The equation for Hooke's Law is F = -kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

Given that the block remains at rest when the spring is stretched a distance of 5.00 cm from its equilibrium length, we can set up the equation as follows:

mg = kx

where m is the mass of the block, g is the acceleration due to gravity, and x is the displacement from the equilibrium position.

Substituting the given values, we have:

(6.00 kg)(9.81 m/s^2) = k(0.05 m)

Solving for k, we get:

k = (6.00 kg)(9.81 m/s^2) / (0.05 m) = 1177.2 N/m

Using the formula for the frequency of a mass-spring system, we can now find the resulting frequency of the block's oscillations:

f = 1 / (2π√(m/k))

Substituting the values, we have:

f = 1 / (2π√(6.00 kg / 1177.2 N/m)) = 1.19 Hz

Answer 2

The frequency of the block's oscillations can be determined by using the equation f = (1/2π) √(k/m), where f is the frequency, k is the spring constant, and m is the mass of the block. By equating the force due to the weight of the block to the force exerted by the spring, we can find the spring constant and substitute it into the frequency equation to find that the resulting frequency is approximately 7.00 Hz.

Step-by-step explanation:

The frequency of the block's oscillations can be determined using the equation:

f = (1/2π) √(k/m)

Where f is the frequency, k is the spring constant, and m is the mass of the block. In this case, the spring constant is unknown. However, we can find it using Hooke's Law:

F = -kx

Where F is the force, k is the spring constant, and x is the displacement from equilibrium. Since the block remains at rest when the spring is stretched by a distance h, we can equate the force due to the weight of the block to the force exerted by the spring:

mg = kx

Solving for k, we have:

k = mg/x

Substituting this value of k back into the equation for the frequency, we get:

f = (1/2π) √(mg/xm)

Simplifying further, we have:

f = (1/2π) √(g/x)

Substituting the given values, we have:

f = (1/2π) √(9.81/0.05) ≈ 7.00 Hz