A small amplitude progressive wave in a stret | vedclass

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(B) Given: Frequency $f = 100 \ Hz$,speed $v = 100 \ cm/s$,and path difference $\Delta x = 2.75 \ cm$.First,we calculate the wavelength $\lambda$ usin

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$11 \frac{\pi}{2}$

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(B) Given: Frequency $f = 100 \ Hz$,speed $v = 100 \ cm/s$,and path difference $\Delta x = 2.75 \ cm$.First,we calculate the wavelength $\lambda$ using the relation $v = f \lambda$:$\lambda = \frac{v}{f} = \frac{100 \ cm/s}{100 \ Hz} = 1 \ cm$.The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Substituting the values:$\Delta \phi = \frac{2 \pi}{1 \ cm} 	imes 2.75 \ cm = 5.5 \pi$.Converting $5.5 \pi$ to a fraction:$5.5 \pi = \frac{11}{2} \pi$ radians.

$A$ small amplitude progressive wave in a stretched string has a speed of $100 \ cm/s$ and a frequency of $100 \ Hz$. The phase difference between two points $2.75 \ cm$ apart on the string,in radians,is

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