To determine which configuration could correctly illustrate standing waves on strings A and B, let's analyze the relationship between the properties of the two strings and their waveforms.

Given information:

1. Length and Tension: Both strings have the same length and the same tension.
2. Mass: String B has 2.25 times the mass of String A.
3. Frequency: Both strings vibrate at the same frequency.

Understanding the Wave Speed Relationship:

The wave speed $v$ on a string is given by the formula: $v = \sqrt{\frac{T}{\mu}}$ where:

• $T$ is the tension in the string (same for both strings),
• $\mu$ is the linear mass density of the string (mass per unit length).

Since String B has 2.25 times the mass of String A, it has 2.25 times the linear mass density as well. Let $\mu_A$ be the mass density of String A, and $\mu_B = 2.25 \mu_A$ for String B.

Substituting into the wave speed formula, we find that: $v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{T}{2.25 \mu_A}} = \frac{v_A}{\sqrt{2.25}} = \frac{v_A}{1.5}$ This shows that the wave speed on String B is slower than on String A by a factor of 1.5.

Relating Wave Speed to Wavelength:

Since both strings vibrate at the same frequency $f$, and wave speed $v$ is related to frequency and wavelength $\lambda$ by $v = f \lambda$, we find: $\lambda_B = \frac{v_B}{f} = \frac{v_A}{1.5 f} = \frac{\lambda_A}{1.5}$ This implies that the wavelength on String B is shorter than that on String A by a factor of 1.5.

Evaluating the Options:

1. Option 1: A has one half wave, B has a full wave

   • Here, the wavelength on String B would be half that of String A, implying $\lambda_B = \frac{\lambda_A}{2}$.
   • This is inconsistent with our calculation ($\lambda_B = \frac{\lambda_A}{1.5}$).

2. Option 2: A is a full wave, B has 1 and a half waves

   • Here, the wavelength on String B is two-thirds that of String A ($\lambda_B = \frac{\lambda_A}{1.5}$), which matches our calculation.
   • This option is possible.

3. Option 3: A has half a wave, B has one and a half waves

   • Here, the wavelength on String B would be one-third that of String A, implying $\lambda_B = \frac{\lambda_A}{3}$.
   • This does not match our calculation.

4. Option 4: A has a wave, B has half a wave

   • This would imply that the wavelength on String B is double that of String A, which does not match our calculation.

5. Option 5: None of the illustrations are correct

   • Since Option 2 is valid, this choice is incorrect.

Conclusion:

The correct answer is: Option 2: A is a full wave, B has 1 and a half waves

Would you like more details or have any questions?

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Related Questions:

1. How does changing the frequency affect the wavelengths on strings A and B?
2. What happens to the wave speed if the tension in both strings is doubled?
3. If the mass of String B were increased to 4 times that of String A, how would the wavelength on B compare to that on A?
4. Why does increasing the mass of a string reduce the wave speed?
5. How would the standing wave patterns change if String A had a higher frequency than String B?

Tip: Remember that wave speed in a string depends on both its tension and its linear mass density, making heavier strings vibrate with shorter wavelengths at the same frequency.