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The water level in a vertical glass tube...

The water level in a vertical glass tube `1.0 m` long can be adjusted to any position in the tube . `A` tuning fork vibrating at `660 H_(Z)` is held just over the open top end of the tube . At what positions of the water level wil ther be in resonance? Speed of sound is `330 m//s` .

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To solve the problem of finding the positions of the water level in a vertical glass tube where resonance occurs when a tuning fork vibrating at 660 Hz is held over the open end, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Values:** - Length of the tube, \( L = 1.0 \, \text{m} \) - Frequency of the tuning fork, \( f = 660 \, \text{Hz} \) - Speed of sound, \( v = 330 \, \text{m/s} \) 2. **Determine the Wavelength (\( \lambda \)):** The wavelength can be calculated using the formula: \[ \lambda = \frac{v}{f} \] Substituting the values: \[ \lambda = \frac{330 \, \text{m/s}}{660 \, \text{Hz}} = 0.5 \, \text{m} \] 3. **Understand the Resonance Condition:** For a tube closed at one end (the bottom is closed by water), the resonance occurs at odd multiples of \( \frac{\lambda}{4} \): \[ L_n = n \frac{\lambda}{4} \quad \text{where } n = 1, 3, 5, \ldots \] 4. **Calculate the Lengths of the Air Column for Resonance:** - For \( n = 1 \): \[ L_1 = 1 \cdot \frac{0.5}{4} = 0.125 \, \text{m} \] - For \( n = 3 \): \[ L_2 = 3 \cdot \frac{0.5}{4} = 0.375 \, \text{m} \] - For \( n = 5 \): \[ L_3 = 5 \cdot \frac{0.5}{4} = 0.625 \, \text{m} \] - For \( n = 7 \): \[ L_4 = 7 \cdot \frac{0.5}{4} = 0.875 \, \text{m} \] 5. **Determine the Water Level Positions:** The water level positions can be found by subtracting the lengths of the air column from the total length of the tube: - For \( L_1 \): \[ \text{Water Level}_1 = 1.0 - 0.125 = 0.875 \, \text{m} \] - For \( L_2 \): \[ \text{Water Level}_2 = 1.0 - 0.375 = 0.625 \, \text{m} \] - For \( L_3 \): \[ \text{Water Level}_3 = 1.0 - 0.625 = 0.375 \, \text{m} \] - For \( L_4 \): \[ \text{Water Level}_4 = 1.0 - 0.875 = 0.125 \, \text{m} \] 6. **Final Water Level Positions:** The positions of the water level where resonance occurs are: - \( 0.125 \, \text{m} \) - \( 0.375 \, \text{m} \) - \( 0.625 \, \text{m} \) - \( 0.875 \, \text{m} \)

To solve the problem of finding the positions of the water level in a vertical glass tube where resonance occurs when a tuning fork vibrating at 660 Hz is held over the open end, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Values:** - Length of the tube, \( L = 1.0 \, \text{m} \) - Frequency of the tuning fork, \( f = 660 \, \text{Hz} \) - Speed of sound, \( v = 330 \, \text{m/s} \) ...
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