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The left-hand end of a long horizontal stretched cord oscillates transversely in SHM with frequency 300 HzHz and amplitude 2.8 cmcm . The cord is under a tension of 180 NN and has a linear density 0.17 kg/mkg/m . At t=0t=0, the end of the cord has an upward displacement of 2.1 cmcm and is falling. Consider the point xx = 0.550 mm on the cord. Determine its maximum acceleration.

To solve this problem, we'll need to use the properties of a wave traveling along a stretched cord, and apply the equations for simple harmonic motion (SHM) and wave mechanics.

### Given Data:
- Frequency \( f = 300 \, \text{Hz} \)
- Amplitude \( A = 2.8 \, \text{cm} = 0.028 \, \text{m} \)
- Tension in the cord \( T = 180 \, \text{N} \)
- Linear density of the cord \( \mu = 0.17 \, \text{kg/m} \)
- At \( t = 0 \), the displacement \( y(0) = 2.1 \, \text{cm} = 0.021 \, \text{m} \)
- The position \( x = 0.550 \, \text{m} \) along the cord

### Step 1: Determine the wave speed \( v \)
The wave speed \( v \) on a string under tension is given by:

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

Substituting the values:

\[
v = \sqrt{\frac{180 \, \text{N}}{0.17 \, \text{kg/m}}} \approx 32.57 \, \text{m/s}
\]

### Step 2: Write the wave equation
The wave equation for a transverse wave traveling along the x-axis can be written as:

\[
y(x, t) = A \sin(kx - \omega t + \phi)
\]

Where:
- \( A \) is the amplitude
- \( k \) is the wave number, \( k = \frac{2\pi}{\lambda} \)
- \( \omega \) is the angular frequency, \( \omega = 2\pi f \)
- \( \phi \) is the phase constant

### Step 3: Determine the angular frequency \( \omega \)
\[
\omega = 2\pi f = 2\pi \times 300 \, \text{Hz} \approx 1885 \, \text{rad/s}
\]

### Step 4: Determine the wave number \( k \)
Using the relationship \( v = \lambda f \):

\[
\lambda = \frac{v}{f} = \frac{32.57 \, \text{m/s}}{300 \, \text{Hz}} \approx 0.1086 \, \text{m}
\]

Thus,

\[
k = \frac{2\pi}{\lambda} \approx \frac{2\pi}{0.1086 \, \text{m}} \approx 57.87 \, \text{rad/m}
\]

### Step 5: Calculate the maximum acceleration
The maximum acceleration for SHM occurs when \( \sin(kx - \omega t + \phi) = 1 \), which corresponds to the maximum value of the sine function. The general formula for acceleration in SHM is:

\[
a_{\text{max}} = \omega^2 A
\]

Substituting the values:

\[
a_{\text{max}} = (1885 \, \text{rad/s})^2 \times 0.028 \, \text{m} \approx 99717 \, \text{m/s}^2
\]

Thus, the maximum acceleration at the point \( x = 0.550 \, \text{m} \) on the cord is approximately **99,717 m/s²**.

Would you like more details or have any questions?

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**Related Questions:**

1. How do different tensions in the cord affect the wave speed?
2. What is the relationship between wave speed and frequency?
3. How does the linear density of the cord influence the wave's characteristics?
4. What is the significance of the phase constant \( \phi \) in the wave equation?
5. How can the wave equation be applied to model other physical phenomena?
6. How does amplitude affect the maximum velocity and acceleration in SHM?
7. How does the initial displacement affect the motion of a wave?
8. What are the effects of damping on a wave traveling through a medium?

**Tip:** Remember that the maximum acceleration in SHM is proportional to both the square of the angular frequency and the amplitude of the oscillation.

Learn how to calculate the maximum acceleration in simple harmonic motion (SHM) on a stretched cord with a detailed solution. Given the frequency, amplitude, tension, and linear density of the cord, determine the maximum acceleration at a specific point on the cord. Suitable for college-level physics students.

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