A string `120 cm` in length sustains a standing wave, with the points of the string at which the displacement amplitude is equal to `3.5 mm` being separated by `15.0cm.` Find the maximum displacement amplitude. To which overtone do these oscillations correspond ?

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understanding the Problem
We have a string of length `L = 120 cm` that sustains a standing wave. The points on the string where the displacement amplitude is `3.5 mm` are separated by `15.0 cm`. We need to find the maximum displacement amplitude and determine the overtone number.

### Step 2: Finding the Wavelength
Since the points of the string where the displacement amplitude is `3.5 mm` are separated by `15 cm`, this distance corresponds to half a wavelength of the standing wave. Therefore, we can express the relationship as:
\[ \frac{\lambda}{2} = 15 \text{ cm} \]
From this, we can find the wavelength:
\[ \lambda = 2 \times 15 \text{ cm} = 30 \text{ cm} \]

### Step 3: Finding the Harmonic Number
The length of the string is related to the wavelength and the harmonic number \( n \) by the formula:
\[ L = n \frac{\lambda}{2} \]
Substituting the values we have:
\[ 120 \text{ cm} = n \frac{30 \text{ cm}}{2} \]
This simplifies to:
\[ 120 = 15n \]
Solving for \( n \):
\[ n = \frac{120}{15} = 8 \]
Thus, the string is vibrating in the 8th harmonic.

### Step 4: Finding the Maximum Displacement Amplitude
The amplitude at a distance \( x \) from the node can be expressed as:
\[ A(x) = A \sin\left(\frac{n \pi x}{L}\right) \]
Given that at \( x = 15 \text{ cm} \) (which is \( \frac{L}{8} \)), the amplitude is \( 3.5 \text{ mm} \):
\[ 3.5 = A \sin\left(\frac{8 \pi \cdot 15}{120}\right) \]
Calculating the sine term:
\[ \frac{8 \pi \cdot 15}{120} = \frac{8 \pi}{8} = \pi \quad \Rightarrow \quad \sin(\pi) = 0 \]
This indicates a misunderstanding; we should instead evaluate at the midpoint between the nodes where the amplitude is maximum.

### Step 5: Correcting the Amplitude Calculation
To find the maximum amplitude, we need to evaluate the amplitude at the midpoint between the nodes:
\[ x = \frac{15 \text{ cm}}{2} = 7.5 \text{ cm} \]
Now, we can use the sine function:
\[ A(7.5) = A \sin\left(\frac{8 \pi \cdot 7.5}{120}\right) \]
Calculating the sine term:
\[ \frac{8 \pi \cdot 7.5}{120} = \frac{8 \pi}{16} = \frac{\pi}{2} \quad \Rightarrow \quad \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus:
\[ A(7.5) = A \]
This means the maximum amplitude \( A \) is given by:
\[ A = \sqrt{2} \times 3.5 \text{ mm} = 5 \text{ mm} \]

### Step 6: Conclusion
The maximum displacement amplitude is \( 5 \text{ mm} \), and the oscillations correspond to the 8th harmonic (or 7th overtone).

### Summary of Results
- Maximum Displacement Amplitude: \( 5 \text{ mm} \)
- Overtone: 7th overtone (8th harmonic)