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A travelling wave on a long stretched st...

A travelling wave on a long stretched string along the positIve x-axis is given by `y = 5mm e^(((T)/(5s) - (x)/(5cm))^2)` . Using this equation answer the following questions.
The velocity of the wave is

A

1m/s

B

5m/s

C

1cm/s

D

1mm/s

Text Solution

AI Generated Solution

The correct Answer is:
To find the velocity of the wave given by the equation \( y = 5 \, \text{mm} \, e^{\left(\frac{T}{5 \, \text{s}} - \frac{x}{5 \, \text{cm}}\right)^2} \), we can follow these steps: ### Step 1: Identify the coefficients The general form of a wave equation can be expressed as: \[ y = A e^{\left(\frac{T}{T_0} - \frac{x}{x_0}\right)^2} \] where \( T_0 \) is the time constant and \( x_0 \) is the spatial constant. From the given equation, we can identify: - Coefficient of \( T \) (time) is \( \frac{1}{5 \, \text{s}} \) - Coefficient of \( x \) (space) is \( \frac{1}{5 \, \text{cm}} \) ### Step 2: Write the formula for wave velocity The velocity \( v \) of a wave can be calculated using the formula: \[ v = -\frac{\text{coefficient of } T}{\text{coefficient of } x} \] ### Step 3: Substitute the coefficients into the formula Now substituting the coefficients we identified: - Coefficient of \( T \) = \( \frac{1}{5 \, \text{s}} \) - Coefficient of \( x \) = \( \frac{1}{5 \, \text{cm}} \) Thus, \[ v = -\frac{\frac{1}{5 \, \text{s}}}{\frac{1}{5 \, \text{cm}}} \] ### Step 4: Simplify the expression Now simplifying the expression: \[ v = -\left(\frac{1}{5 \, \text{s}} \cdot \frac{5 \, \text{cm}}{1}\right) \] \[ v = -1 \, \text{cm/s} \] Since we are interested in the magnitude of the velocity, we take the absolute value: \[ v = 1 \, \text{cm/s} \] ### Final Answer The velocity of the wave is \( 1 \, \text{cm/s} \). ---
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