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the equation of a wave on a string of li...

the equation of a wave on a string of linear mass density `0.04 kgm^(-1)` is given by
`y = 0.02(m) sin[2pi((t)/(0.04(s)) -(x)/(0.50(m)))]`.
Then tension in the string is

A

(a) `4.0 N`

B

(b) `12.5 N`

C

( c ) `0.5 N`

D

(d) `6.25 N`

Text Solution

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The correct Answer is:
To find the tension in the string given the wave equation, we can follow these steps: ### Step 1: Identify the parameters from the wave equation The wave equation is given as: \[ y = 0.02 \, \text{m} \, \sin\left[2\pi\left(\frac{t}{0.04 \, \text{s}} - \frac{x}{0.50 \, \text{m}}\right)\right] \] From this equation, we can identify: - Amplitude \( A = 0.02 \, \text{m} \) - Angular frequency \( \omega = 2\pi \left(\frac{1}{0.04}\right) \) - Wave number \( k = 2\pi \left(\frac{1}{0.50}\right) \) ### Step 2: Calculate angular frequency \( \omega \) and wave number \( k \) Calculating \( \omega \): \[ \omega = 2\pi \left(\frac{1}{0.04}\right) = \frac{2\pi}{0.04} = 50\pi \, \text{rad/s} \] Calculating \( k \): \[ k = 2\pi \left(\frac{1}{0.50}\right) = \frac{2\pi}{0.50} = 4\pi \, \text{rad/m} \] ### Step 3: Calculate the wave velocity \( v \) The wave velocity \( v \) can be calculated using the relationship: \[ v = \frac{\omega}{k} \] Substituting the values: \[ v = \frac{50\pi}{4\pi} = \frac{50}{4} = 12.5 \, \text{m/s} \] ### Step 4: Use the formula for tension in the string The tension \( T \) in the string can be calculated using the formula: \[ v = \sqrt{\frac{T}{\mu}} \] Where \( \mu \) is the linear mass density. Rearranging gives: \[ T = \mu v^2 \] ### Step 5: Substitute the values to find tension Given \( \mu = 0.04 \, \text{kg/m} \) and \( v = 12.5 \, \text{m/s} \): \[ T = 0.04 \times (12.5)^2 \] Calculating \( (12.5)^2 \): \[ (12.5)^2 = 156.25 \] Now substituting back: \[ T = 0.04 \times 156.25 = 6.25 \, \text{N} \] ### Final Answer The tension in the string is \( T = 6.25 \, \text{N} \). ---

To find the tension in the string given the wave equation, we can follow these steps: ### Step 1: Identify the parameters from the wave equation The wave equation is given as: \[ y = 0.02 \, \text{m} \, \sin\left[2\pi\left(\frac{t}{0.04 \, \text{s}} - \frac{x}{0.50 \, \text{m}}\right)\right] \] From this equation, we can identify: - Amplitude \( A = 0.02 \, \text{m} \) ...
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