PHYSICS WAVE MOTION: In an experiment on standing waves, a string 31 cm long is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of 60 Hz.The mass of the string is 0.011 kg. What tension must the string be under(weights are attached to other end) if it is to oscillate in four loops

To solve for the tension in the string that will allow it to oscillate in four loops (four antinodes), we need to use the wave equation for standing waves on a string.

### Key formulas:
1. **Wave velocity on a string:**
   \[
   v = \sqrt{\frac{T}{\mu}}
   \]
   where:
   - \( v \) is the wave speed,
   - \( T \) is the tension in the string,
   - \( \mu \) is the linear mass density (mass per unit length) of the string.

2. **Linear mass density:**
   \[
   \mu = \frac{m}{L}
   \]
   where:
   - \( m \) is the mass of the string,
   - \( L \) is the length of the string.

3. **Frequency of the standing wave:**
   The frequency of a standing wave is related to the number of loops (antinodes) and the length of the string by the equation:
   \[
   f = \frac{n v}{2 L}
   \]
   where:
   - \( f \) is the frequency of the wave,
   - \( n \) is the number of loops (antinodes),
   - \( v \) is the wave speed,
   - \( L \) is the length of the string.

Now, let's proceed step by step.

### Step 1: Calculate the linear mass density (\( \mu \))
Given:
- Mass of the string, \( m = 0.011 \, \text{kg} \),
- Length of the string, \( L = 31 \, \text{cm} = 0.31 \, \text{m} \).

\[
\mu = \frac{m}{L} = \frac{0.011 \, \text{kg}}{0.31 \, \text{m}} = 0.03548 \, \text{kg/m}
\]

### Step 2: Use the standing wave frequency equation
For four loops, \( n = 4 \), and the frequency is \( f = 60 \, \text{Hz} \). We can rearrange the frequency equation to solve for \( v \) (wave speed):
\[
v = \frac{2 L f}{n}
\]
Substitute the known values:
\[
v = \frac{2 \times 0.31 \, \text{m} \times 60 \, \text{Hz}}{4} = 9.3 \, \text{m/s}
\]

### Step 3: Solve for tension (\( T \))
Now, use the wave velocity equation \( v = \sqrt{\frac{T}{\mu}} \) to solve for tension \( T \):
\[
T = v^2 \mu
\]
Substitute the values:
\[
T = (9.3 \, \text{m/s})^2 \times 0.03548 \, \text{kg/m} = 3.07 \, \text{N}
\]

### Final Answer:
The tension required in the string is approximately **3.07 N**.

Would you like further details or clarifications on any part of this solution?

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Here are 5 related questions to further explore this topic:
1. How does the number of loops affect the required tension in a string for standing waves?
2. What happens to the frequency if the tension in the string is doubled?
3. How would the wave speed change if the mass of the string was doubled?
4. How is the wavelength related to the length of the string for standing waves?
5. What is the effect of changing the length of the string on the wave speed?

**Tip:** The number of loops in a standing wave is directly related to the harmonic mode of vibration. The more loops, the higher the harmonic number.

PHYSICS WAVE MOTION: In an experiment on standing waves, a string 31 cm long is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of 60 Hz. The mass of the string is 0.011 kg. What tension must the string be under (weights are attached to other end) if it is to oscillate in four loops?

Find the tension needed in a string to produce standing waves in four loops with a frequency of 60 Hz. This problem explores key concepts in wave motion, including linear mass density and wave velocity. Suitable for high school students studying physics.

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