In which case is the power being delivered by a given progressive sinusoidal wave on a given string is doubled? Material of the string is unchanged.

Options

A. The wave amplitude is doubled (keeping the frequency the same).
B. The wave frequency is cut in half (keeping the amplitude the same).
C. Tension in the string is made four times the initial value (keeping its linear density, amplitude and frequency the same).
D. The diameter of the string is doubled (keeping the tension amplitude and frequency the same).

Answer

To determine in which case the power delivered by a given progressive sinusoidal wave on a string is doubled, we need to analyze the relationship between power, amplitude, frequency, tension, and linear density of the string. The power \( P \) transmitted by a wave on a string can be expressed as:

\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]

Where:
- \( \mu \) is the linear density of the string (mass per unit length),
- \( \omega \) is the angular frequency of the wave,
- \( A \) is the amplitude of the wave,
- \( v \) is the wave speed on the string.

The wave speed \( v \) can be expressed in terms of tension \( T \) and linear density \( \mu \):

\[ v = \sqrt{\frac{T}{\mu}} \]

### Step-by-Step Solution:

1. **Identify the parameters affecting power**:
The power delivered by the wave depends on amplitude \( A \), frequency (via angular frequency \( \omega \)), tension \( T \), and linear density \( \mu \).

2. **Case 1 - Doubling the amplitude**:
If the amplitude \( A \) is doubled (i.e., \( A = 2A_0 \)), the power becomes:
\[ P' = \frac{1}{2} \mu \omega^2 (2A_0)^2 v = \frac{1}{2} \mu \omega^2 (4A_0^2) v = 4P_0 \]
This quadruples the power, not doubles it.

3. **Case 2 - Halving the frequency**:
If the frequency is cut in half, then \( \omega = \frac{1}{2} \omega_0 \):
\[ P' = \frac{1}{2} \mu \left(\frac{1}{2} \omega_0\right)^2 A_0^2 v = \frac{1}{2} \mu \frac{1}{4} \omega_0^2 A_0^2 v = \frac{1}{8} P_0 \]
This reduces the power, not doubles it.

4. **Case 3 - Increasing tension**:
If the tension \( T \) is made four times the initial value (i.e., \( T = 4T_0 \)), the wave speed becomes:
\[ v = \sqrt{\frac{4T_0}{\mu}} = 2\sqrt{\frac{T_0}{\mu}} = 2v_0 \]
Thus, the power becomes:
\[ P' = \frac{1}{2} \mu \omega^2 A_0^2 (2v_0) = 2P_0 \]
This doubles the power.

5. **Case 4 - Doubling the diameter**:
If the diameter of the string is doubled, the linear density \( \mu \) increases (since \( \mu \) is proportional to the cross-sectional area, which is proportional to the square of the diameter). The new linear density becomes:
\[ \mu' = 4\mu_0 \]
The power becomes:
\[ P' = \frac{1}{2} (4\mu_0) \omega^2 A_0^2 v \]
This increases the power by a factor of 4, not doubles it.

### Conclusion:
The only case in which the power delivered by the wave is doubled is when the tension in the string is made four times its initial value while keeping the linear density, amplitude, and frequency the same.