String 1 in the figure has linear density 2.80 g/m
and string 2 has linear density 3.30 g/m
. A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. She wants both pulses to reach the ends of the strings simultaneously. What should the string length  L1
be?

To solve this problem, we need to ensure that the pulses sent by the student reach the ends of both strings simultaneously. This involves finding the appropriate length \( L_1 \) for String 1, given the linear densities of the strings and the condition for simultaneous arrival of the pulses.

### Step-by-Step Solution:

1. **Understand the Given Information:**
   - Linear density of String 1 (\( \mu_1 \)): 2.80 g/m
   - Linear density of String 2 (\( \mu_2 \)): 3.30 g/m
   - We need to find the length \( L_1 \) of String 1 such that the time it takes for the pulse to travel through String 1 is the same as the time it takes to travel through String 2.

2. **Convert Linear Densities to Consistent Units:**
   - Convert the linear densities from grams per meter to kilograms per meter (since 1 g = 0.001 kg):
   \[
   \mu_1 = 2.80 \, \text{g/m} = 0.0028 \, \text{kg/m}
   \]
   \[
   \mu_2 = 3.30 \, \text{g/m} = 0.0033 \, \text{kg/m}
   \]

3. **Use the Wave Speed Formula:**
   The speed \( v \) of a wave on a string is given by:
   \[
   v = \sqrt{\frac{T}{\mu}}
   \]
   where \( T \) is the tension in the string, and \( \mu \) is the linear density.

   Since the strings are attached to the same knot, the tension \( T \) in both strings is the same.

4. **Find the Speeds of Pulses on Each String:**
   - Speed on String 1 (\( v_1 \)):
   \[
   v_1 = \sqrt{\frac{T}{\mu_1}}
   \]
   - Speed on String 2 (\( v_2 \)):
   \[
   v_2 = \sqrt{\frac{T}{\mu_2}}
   \]

5. **Set Up the Condition for Simultaneous Arrival:**
   To have both pulses arrive simultaneously, the travel time for each pulse must be the same. The travel time \( t \) for a pulse is given by:
   \[
   t = \frac{L}{v}
   \]
   Therefore, for the pulses to reach the ends simultaneously:
   \[
   \frac{L_1}{v_1} = \frac{L_2}{v_2}
   \]

6. **Substitute the Speeds and Solve for \( L_1 \):**
   Substitute \( v_1 \) and \( v_2 \) into the equation:
   \[
   \frac{L_1}{\sqrt{\frac{T}{\mu_1}}} = \frac{L_2}{\sqrt{\frac{T}{\mu_2}}}
   \]
   Simplify by multiplying both sides by \( \sqrt{\frac{T}{\mu_1}} \sqrt{\frac{T}{\mu_2}} \) (to eliminate \( T \)):
   \[
   L_1 \sqrt{\mu_2} = L_2 \sqrt{\mu_1}
   \]
   Solve for \( L_1 \):
   \[
   L_1 = L_2 \frac{\sqrt{\mu_1}}{\sqrt{\mu_2}}
   \]

7. **Plug in the Values for \( \mu_1 \) and \( \mu_2 \):**
   \[
   L_1 = L_2 \frac{\sqrt{0.0028}}{\sqrt{0.0033}}
   \]
   Calculate the square roots:
   \[
   L_1 = L_2 \frac{\sqrt{0.0028}}{\sqrt{0.0033}} = L_2 \frac{0.052915}{0.057445}
   \]
   \[
   L_1 \approx L_2 \times 0.9214
   \]

Therefore, the length of String 1 \( L_1 \) should be approximately 92.14% of the length of String 2 \( L_2 \).

### Conclusion

To make sure both pulses reach the ends of the strings simultaneously, \( L_1 \) should be 0.9214 times \( L_2 \).

Would you like further details on any part of the solution? Here are some related questions you might find interesting:

1. What happens to the pulse speeds if the tension in the strings is increased?
2. How does changing the linear density affect the wave speed?
3. Can we calculate the exact time it takes for a pulse to travel down the string?
4. What are some practical applications of this wave propagation principle?
5. How would the result change if the strings had different tensions?

**Tip:** Remember that wave speed on a string depends on both the tension and the linear density; manipulating either can change the speed at which waves travel through the string.

Learn how to calculate the length of String 1 to ensure simultaneous arrival of wave pulses with String 2. This problem involves understanding wave speed, linear density, and wave propagation principles in physics. Suitable for high school physics students.

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