Question

frequency and wavelength relationships example
calculate the frequency given the information in each table
calculate the wavelength given the information in each table
in your own words, explain how frequency and wavelength are related

Explanation:

Step1: Recall the wave - speed formula

The wave - speed formula is $v = f\lambda$, where $v$ is the wave speed, $f$ is the frequency, and $\lambda$ is the wavelength. Assuming the wave speed is constant (which is the case for waves in a uniform medium), we have $f_1\lambda_1=f_2\lambda_2$.

Step2: Solve for the unknown frequency in the first table

Given $\lambda_1 = 25m$, $f_1 = 15Hz$, and $\lambda_2 = 5m$. Using $f_1\lambda_1=f_2\lambda_2$, we can solve for $f_2$:
\[f_2=\frac{f_1\lambda_1}{\lambda_2}=\frac{15\times25}{5}=75Hz\]

Step3: Solve for the unknown frequency in the second table

Given $\lambda_1 = 18m$, $f_1 = 12Hz$, and $\lambda_2 = 4m$. Using $f_1\lambda_1=f_2\lambda_2$, we get $f_2=\frac{f_1\lambda_1}{\lambda_2}=\frac{12\times18}{4}=54Hz$

Step4: Solve for the unknown wavelength in the third table

Given $f_1 = 20Hz$, $\lambda_1 = 15m$, and $f_2 = 300Hz$. Using $f_1\lambda_1=f_2\lambda_2$, we solve for $\lambda_2$:
\[\lambda_2=\frac{f_1\lambda_1}{f_2}=\frac{20\times15}{300}=1m\]

Step5: Solve for the unknown frequency in the fourth table

Given $\lambda_1 = 80m$, $f_1 = 20Hz$, and $\lambda_2 = 24m$. Using $f_1\lambda_1=f_2\lambda_2$, we have $f_2=\frac{f_1\lambda_1}{\lambda_2}=\frac{20\times80}{24}=\frac{200}{3}\approx66.67Hz$

Step6: Solve for the unknown wavelength in the fifth table

Given $f_1 = 75Hz$, $\lambda_1 = 200m$, and $f_2 = 480Hz$. Using $f_1\lambda_1=f_2\lambda_2$, we get $\lambda_2=\frac{f_1\lambda_1}{f_2}=\frac{75\times200}{480}=\frac{125}{4}=31.25m$

Step7: Solve for the unknown frequency in the sixth table

Given $\lambda_1 = 2m$, $f_1 = 150Hz$, and $\lambda_2 = 10m$. Using $f_1\lambda_1=f_2\lambda_2$, we have $f_2=\frac{f_1\lambda_1}{\lambda_2}=\frac{150\times2}{10}=30Hz$

Step8: Solve for the unknown wavelength in the seventh table

Given $f_1 = 582Hz$, $\lambda_1 = 71m$, and $f_2 = 194Hz$. Using $f_1\lambda_1=f_2\lambda_2$, we get $\lambda_2=\frac{f_1\lambda_1}{f_2}=\frac{582\times71}{194}=213m$

Step9: Solve for the unknown frequency in the eighth table

Given $\lambda_1 = 27m$, $f_1 = 32Hz$, and $\lambda_2 = 18m$. Using $f_1\lambda_1=f_2\lambda_2$, we have $f_2=\frac{f_1\lambda_1}{\lambda_2}=\frac{32\times27}{18}=48Hz$

Step10: Solve for the unknown wavelength in the ninth table

Given $f_1 = 7Hz$, $\lambda_1 = 48m$, and $f_2 = 168Hz$. Using $f_1\lambda_1=f_2\lambda_2$, we get $\lambda_2=\frac{f_1\lambda_1}{f_2}=\frac{7\times48}{168}=2m$

The relationship between frequency and wavelength: When the wave speed is constant, the frequency and the wavelength of a wave are inversely proportional. That is, as the wavelength decreases, the frequency increases, and vice - versa.

Answer:

First table: 75 Hz
Second table: 54 Hz
Third table: 1 m
Fourth table: $\frac{200}{3}Hz\approx66.67Hz$
Fifth table: 31.25 m
Sixth table: 30 Hz
Seventh table: 213 m
Eighth table: 48 Hz
Ninth table: 2 m
Relationship: Inversely proportional when wave - speed is constant.