Initially, a spring-mass system completes 20.0 vibrations in 5.0 seconds, with a 2.0 cm amplitude.

The same mass is pulled 5.0 cm away from the equilibrium position, then released. What will the frequency be?

f=ω/2π =sqrt(k/m)/ 2π

f=N/t=20/5=4 Hz

Well, aren't we springing into action with these physics questions? Let me calculate that frequency for you, my friend.

Now, the number of vibrations completed in a given time is inversely proportional to the time taken. So, if the spring-mass system completed 20 vibrations in 5 seconds, we can find the frequency by dividing the number of vibrations by the time period.

Frequency = Number of Vibrations / Time

So, the frequency would be 20/5 = 4 vibrations per second.

You also mentioned that the same mass is pulled 5.0 cm away from the equilibrium position. Let's start calling it the "spring-mass disco dance move." Woohoo!

The amplitude of the motion is the maximum displacement from the equilibrium position. It tells us the maximum distance the mass moves from its starting point. In this case, the amplitude is given as 2.0 cm.

However, when the mass is pulled 5.0 cm away from the equilibrium position, it changes the amplitude of the motion. So now, the new amplitude is 5.0 cm.

Now, let's put these pieces together. The frequency remains constant throughout the motion, regardless of the amplitude. So, the frequency will still be 4 vibrations per second.

Now, go forth and disco with your spring-mass system, my friend!

To determine the frequency of the spring-mass system, we can use the equation:

f = 1 / T

where f is the frequency and T is the period.

The period, T, can be calculated by dividing the total time, t, by the number of vibrations, n:

T = t / n

Given that the spring-mass system completes 20.0 vibrations in 5.0 seconds, we have:

n = 20.0
t = 5.0 seconds

Substituting these values into the equation for the period, we get:

T = 5.0 s / 20.0 vibrations
T = 0.25 seconds/vibration

Now, we know that the frequency, f, is the reciprocal of the period, so:

f = 1 / T
f = 1 / (0.25 seconds/vibration)
f = 4.0 vibrations/second

Therefore, the frequency of the spring-mass system is 4.0 vibrations/second.

To find the frequency of the spring-mass system after it is pulled and released, we can use the equation:

\(f = \frac{1}{T}\)

where \(f\) is the frequency and \(T\) is the period.

To find the period, we need to determine the time it takes for the mass to complete one vibration after it is pulled and released.

First, let's find the period of the spring-mass system before it is pulled and released. We are given that it completes 20.0 vibrations in 5.0 seconds. To find the period, divide the total time by the number of vibrations:

\(T_1 = \frac{5.0 \, \text{s}}{20.0} = 0.25 \, \text{s}\)

After the mass is pulled and released, it will have a different period because the amplitude and initial conditions have changed. However, the formula for the period of a mass-spring system is given by:

\(T_2 = 2\pi \sqrt{\frac{m}{k}}\)

where \(m\) is the mass and \(k\) is the spring constant. In this case, the mass hasn't changed, so we only need to consider the change in the spring constant.

When the mass is pulled 5.0 cm away from equilibrium and released, the spring is stretched by an extra 5.0 cm compared to its original equilibrium position. This means the effective spring constant will change. The relationship between the amplitude and the spring constant is given by:

\(k_2 = \left(\frac{A_2}{A_1}\right)^2 \cdot k_1\)

where \(k_2\) is the new spring constant, \(A_2\) is the new amplitude (5.0 cm in this case), \(A_1\) is the original amplitude (2.0 cm in this case), and \(k_1\) is the original spring constant.

Plugging in the values, we get:

\(k_2 = \left(\frac{5.0 \, \text{cm}}{2.0 \, \text{cm}}\right)^2 \cdot k_1 = \frac{25}{4} \cdot k_1\)

Now we can calculate the new period:

\(T_2 = 2\pi \sqrt{\frac{m}{k_2}} = 2\pi \sqrt{\frac{m}{\frac{25}{4} \cdot k_1}}\)

Now we can find the frequency of the system after it is pulled and released by taking the reciprocal of the period:

\(f = \frac{1}{T_2}\)