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A transverse periodic wave ona strin wit...

A transverse periodic wave ona strin with a linear mass density of 0.200kg/m is described by the following equations
`y=0.05sin(420t-21.0x)`
where x and y in metres and t is in seconds. Tension in the string is

A

32N

B

42N

C

66N

D

80N

Text Solution

AI Generated Solution

The correct Answer is:
To find the tension in the string described by the wave equation \( y = 0.05 \sin(420t - 21.0x) \), we can follow these steps: ### Step 1: Identify the angular frequency and wave number From the wave equation \( y = 0.05 \sin(420t - 21.0x) \), we can identify: - Angular frequency \( \omega = 420 \, \text{rad/s} \) - Wave number \( k = 21.0 \, \text{rad/m} \) ### Step 2: Calculate the wave speed The wave speed \( v \) can be calculated using the formula: \[ v = \frac{\omega}{k} \] Substituting the values: \[ v = \frac{420}{21.0} = 20 \, \text{m/s} \] ### Step 3: Calculate the wavelength The wavelength \( \lambda \) can be calculated using the wave number: \[ \lambda = \frac{2\pi}{k} \] Substituting the value of \( k \): \[ \lambda = \frac{2\pi}{21.0} \approx 0.299 \, \text{m} \] ### Step 4: Relate wave speed to tension and linear mass density We know that the wave speed \( v \) is also related to the tension \( T \) and the linear mass density \( \mu \) by the formula: \[ v = \sqrt{\frac{T}{\mu}} \] Squaring both sides gives: \[ v^2 = \frac{T}{\mu} \] ### Step 5: Solve for tension Rearranging the equation to solve for tension \( T \): \[ T = \mu v^2 \] Substituting the known values (\( \mu = 0.200 \, \text{kg/m} \) and \( v = 20 \, \text{m/s} \)): \[ T = 0.200 \times (20)^2 = 0.200 \times 400 = 80 \, \text{N} \] ### Final Answer The tension in the string is \( T = 80 \, \text{N} \). ---
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