Question

1. set the frequency so it is now at 3.00 hz. keep the amplitude at 1.00 cm.
2. restart the wave. play the oscillation, then pause it.
3. move the horizontal ruler to measure the wavelength. wavelength #5 =
4. summarize your data in the chart below.

data #	frequency (hz)	wavelength (cm)
#2	1.00 hz	5.8
#3	1.50 hz	4.2
#4	2.00 hz	3
#5	3.00 hz	2.4

1. why did the summary chart not include any information about the amplitude?
2. did changing the frequency affect the wavelength?
3. as the frequency increased, what happened to the wavelength? (look at your data)
4. what happens to the wavelength of a wave if the frequency is doubled?
5. what happens to the wavelength of a wave if the frequency is tripled?

Explanation:

Step1: Recall wave - speed formula

The speed of a wave is given by $v = f\lambda$, where $v$ is the wave - speed, $f$ is the frequency, and $\lambda$ is the wavelength. When the wave - speed is constant (which is likely the case in this experiment as no information about changing the medium is given), frequency and wavelength are inversely proportional.

Step2: Analyze question 20

The summary chart focuses on the relationship between frequency and wavelength. Since the amplitude was kept constant ($1.00$ cm) throughout the experiment, it is not relevant to the study of how frequency affects wavelength. So, it is not included in the chart.

Step3: Analyze question 21

From the data in the chart, as the frequency increases from $1.00$ Hz to $3.00$ Hz, the wavelength decreases from $6$ cm to $2.4$ cm. So, changing the frequency does affect the wavelength.

Step4: Analyze question 22

As the frequency increases, according to the inverse - proportionality relationship $v = f\lambda$ (with $v$ constant), the wavelength decreases. This is consistent with the data in the chart.

Step5: Analyze question 23

If the frequency is doubled, since $v = f\lambda$ and $v$ is constant, the wavelength is halved. For example, if the initial frequency is $f_1$ and wavelength is $\lambda_1$, and the new frequency $f_2 = 2f_1$, then $v=f_1\lambda_1=f_2\lambda_2 = 2f_1\lambda_2$, so $\lambda_2=\frac{\lambda_1}{2}$.

Step6: Analyze question 24

If the frequency is tripled, since $v = f\lambda$ and $v$ is constant, the wavelength is reduced to one - third of its original value. If the initial frequency is $f_1$ and wavelength is $\lambda_1$, and the new frequency $f_3 = 3f_1$, then $v=f_1\lambda_1=f_3\lambda_3 = 3f_1\lambda_3$, so $\lambda_3=\frac{\lambda_1}{3}$.

Answer:

1. The amplitude was kept constant throughout the experiment, so it is not relevant to the relationship between frequency and wavelength being studied.
2. Yes
3. It decreases.
4. It is halved.
5. It is reduced to one - third of its original value.