A spring stretches by 0.018 m when a 3.6-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 2.9 Hz?

You do not need to find the difference. The second to last step is the answer.

k=mg/x=3.6•9.8/0.018 =1960

f=2.9 Hz => ω=2πf=2π•2.9 = 18.22 rad/s
ω²=k/M => M=k/ ω²=1960/(18.22)² = 5.9 kg
Δm=M-m=5.9 – 3.6 – 2.3 kg

To find the mass that should be attached to the spring, we can use Hooke's Law and the formula for the frequency of vibrations of a mass-spring system.

1. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula is given by:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

2. From the given information, we know that the spring stretches by 0.018 m. This is equivalent to the displacement x.

Therefore, x = -0.018 m (Negative sign because the displacement is downwards in this case)

3. We can rearrange Hooke's Law to solve for the spring constant, k:

k = -F / x

The force F can be calculated using Newton's second law:

F = m * g

where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting F and x into the equation for k:

k = -(m * g) / x

4. The frequency of vibrations of a mass-spring system is given by:

f = (1 / 2π) * sqrt(k / m)

Rearranging the equation to solve for m:

m = (k / (4π^2 * f^2))

5. Substitute the value of k from step 3 into the equation in step 4:

m = (-(m * g) / x) / (4π^2 * f^2)
(m * x) / (4π^2 * f^2) = -m * g
mx / (4π^2 * f^2) = -m * g

6. Divide both sides of the equation by -g:

mx / (4π^2 * f^2 * -g) = -m
(mx) / (4π^2 * f^2 * g) = m

7. Finally, substitute the known values into the equation from step 6:

m = (3.6 * 0.018) / (4π^2 * (2.9)^2 * 9.8)

8. Calculate the value of m:

m ≈ 0.072 kg

Therefore, approximately 0.072 kg of mass should be attached to the spring to achieve a frequency of vibration of 2.9 Hz.

To solve this problem, we need to use Hooke's Law and the formula for the frequency of vibration of a mass-spring system.

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from the equilibrium position. Mathematically, it can be expressed as:

F = -kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement.

The frequency of vibration of a mass-spring system can be calculated using the formula:

f = 1 / (2π) * √(k / m)

where f is the frequency of vibration, k is the spring constant, and m is the mass attached to the spring.

In this problem, we are given the displacement of the spring (0.018 m) and the frequency of vibration (2.9 Hz). We can use this information to find the spring constant.

First, we need to rearrange the second equation to solve for k:

k = (2πf)^2 * m

Next, we substitute the given values:

k = (2π * 2.9)^2 * 3.6

Now, we can calculate the spring constant:

k ≈ 40.741 N/m

Finally, we can calculate the mass that should be attached to the spring using the rearranged formula:

m = (2πf)^2 / k

m = (2π * 2.9)^2 / 40.741

m ≈ 0.0745 kg

Therefore, approximately 0.0745 kg of mass should be attached to the spring for its frequency of vibration to be 2.9 Hz.