Two different types of rope with different properties are fused together to make one long rope. When a wave transfers from the first type of rope to the second type of rope, the wavelength becomes one-fourth of what it was before the transfer. What is true about the speed of the wave?(1 point) Responses The speed of the wave becomes four times its original speed after the transfer. The speed of the wave becomes four times its original speed after the transfer. The speed of the wave becomes double its original speed after the transfer. The speed of the wave becomes double its original speed after the transfer. The speed of the wave becomes one-fourth its original speed after the transfer. The speed of the wave becomes one-fourth its original speed after the transfer. The speed of the wave remains the same after the transfer.

1 answer

To analyze what happens when a wave transitions from one medium (the first type of rope) to another (the second type of rope) with a change in wavelength, we can use the wave speed equation:

\[ v = f \lambda \]

where:

  • \( v \) is the wave speed,
  • \( f \) is the frequency,
  • \( \lambda \) is the wavelength.

When the wave transitions to the second medium and the wavelength becomes one-fourth of what it was before, we have:

\[ \lambda_2 = \frac{1}{4} \lambda_1 \]

However, the frequency remains constant during the transition between materials because of the continuity of the wave. Thus, we have:

\[ f_1 = f_2 \]

We can express the wave speeds in the two mediums as follows:

\[ v_1 = f_1 \lambda_1 \] \[ v_2 = f_2 \lambda_2 \]

Substituting \( f_2 = f_1 \) and \( \lambda_2 = \frac{1}{4} \lambda_1 \) into the second equation gives:

\[ v_2 = f_1 \left(\frac{1}{4} \lambda_1\right) = \frac{1}{4} f_1 \lambda_1 = \frac{1}{4} v_1 \]

This shows that the wave speed in the second rope (medium) is one-fourth of the wave speed in the first rope. Hence, the correct response is:

The speed of the wave becomes one-fourth its original speed after the transfer.