Question

1. how did the frequency of the wave change as the wavelength decreased from trial 1 to trial 2?

Explanation:

To answer this, we use the wave equation \( v = f\lambda \) (where \( v \) is wave speed, \( f \) is frequency, \( \lambda \) is wavelength). Assuming the wave speed (\( v \)) remains constant (e.g., in a given medium for light or sound under constant conditions), we can rearrange the formula to \( f=\frac{v}{\lambda} \).

Step 1: Analyze the relationship

From \( f=\frac{v}{\lambda} \), frequency (\( f \)) and wavelength (\( \lambda \)) are inversely proportional when speed (\( v \)) is constant. This means if one quantity increases, the other decreases, and vice versa, as long as \( v \) doesn't change.

Step 2: Apply to the problem

The question states the wavelength (\( \lambda \)) decreased from Trial 1 to Trial 2. Since \( f \) and \( \lambda \) are inversely proportional (with constant \( v \)), when \( \lambda \) decreases, \( f \) must increase (because \( v \) is in the numerator and \( \lambda \) in the denominator—reducing the denominator with a constant numerator makes the whole fraction larger).

Answer:

As the wavelength decreased from Trial 1 to Trial 2, the frequency of the wave increased (assuming the wave speed remained constant, based on the inverse relationship between frequency and wavelength from the wave equation \( v = f\lambda \)).