The function y(x,t) =(15.0 cm) `cos(pi x-15pit),` with x in meters and t in seconds, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement y= +6.00 cm?

Answer

To find the transverse speed of a point on the string when its displacement is \( y = +6.00 \, \text{cm} \), we will follow these steps:

### Step 1: Write down the wave function
The given wave function is:
\[ y(x,t) = 15.0 \, \text{cm} \cdot \cos(\pi x - 15\pi t) \]

### Step 2: Set the displacement equal to 6 cm
We need to find the time \( t \) when the displacement \( y \) is \( +6.0 \, \text{cm} \):
\[ 6.0 \, \text{cm} = 15.0 \, \text{cm} \cdot \cos(\pi x - 15\pi t) \]

### Step 3: Solve for the cosine term
Dividing both sides by \( 15.0 \, \text{cm} \):
\[ \frac{6.0 \, \text{cm}}{15.0 \, \text{cm}} = \cos(\pi x - 15\pi t) \]
\[ 0.4 = \cos(\pi x - 15\pi t) \]

### Step 4: Find the angle
Now, we find the angle whose cosine is \( 0.4 \):
\[ \pi x - 15\pi t = \cos^{-1}(0.4) \]

Calculating \( \cos^{-1}(0.4) \):
\[ \cos^{-1}(0.4) \approx 1.1593 \, \text{radians} \quad (\text{or } 66.42^\circ) \]

### Step 5: Differentiate the wave function with respect to time
To find the transverse speed, we need to differentiate \( y(x,t) \) with respect to time \( t \):
\[ \frac{\partial y}{\partial t} = \frac{\partial}{\partial t} [15.0 \, \text{cm} \cdot \cos(\pi x - 15\pi t)] \]
Using the chain rule:
\[ \frac{\partial y}{\partial t} = 15.0 \, \text{cm} \cdot (-\sin(\pi x - 15\pi t)) \cdot (-15\pi) \]
\[ = 225\pi \, \text{cm} \cdot \sin(\pi x - 15\pi t) \]

### Step 6: Substitute the angle into the derivative
Now substitute \( \pi x - 15\pi t = \cos^{-1}(0.4) \):
\[ \frac{\partial y}{\partial t} = 225\pi \, \text{cm} \cdot \sin(1.1593) \]

Calculating \( \sin(1.1593) \):
\[ \sin(1.1593) \approx 0.9397 \]

### Step 7: Calculate the transverse speed
Now, substituting this value back in:
\[ \frac{\partial y}{\partial t} \approx 225\pi \cdot 0.9397 \, \text{cm/s} \]
Calculating this gives:
\[ \frac{\partial y}{\partial t} \approx 225 \cdot 3.1416 \cdot 0.9397 \approx 663.5 \, \text{cm/s} \approx 6.635 \, \text{m/s} \]

### Final Answer
The transverse speed for a point on the string when the displacement is \( +6.00 \, \text{cm} \) is approximately \( 6.64 \, \text{m/s} \).

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