A triangle wave pulse on a taut string travels in positive x–direction with speed v. the tension in the string is F, and linear mass density of string is $μ$ . At t = 0 the shape of pulse is given by
$y (x, 0) = \{\begin{array}{cc} 0 & if x < - L \\ \frac{h (L + x)}{L} & for - L < x < 0 \\ \frac{h (L - x)}{L} & for 0 < x < L \\ 0 & for x > L \end{array}$
Choose the INCORRECT statement(s). (h << L)

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Magnitude of instantaneous power is zero for $- L < (x - v t) < - L$

Magnitude of instantaneous power is $Fv {(\frac{h}{L})}^{2}$ for (x – vt) > L

Magnitude of instantaneous power is $Fv {(\frac{h}{L})}^{2}$ for 0 < (x – vt) < L

Magnitude of instantaneous power is $Fv {(\frac{h}{L})}^{2}$ for (x – vt) < – L

answer is A, B, C.

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Detailed Solution

$Wave moves in positive x direction ∴ replace (x) by (x - vt) to get y (x, t)$

$y (x, t) = \{\begin{array}{cc} 0 & if (x - vt) < - L \\ \frac{h (L + (x - vt))}{L} & for - L < (x - vt) < 0 \\ \frac{h (L - (x - vt))}{L} & for 0 < (x - vt) < L \\ 0 & for (x - vt) > L \end{array}$

$P o w e r = P (x, t) = - F (\frac{\partial y}{\partial x}) (\frac{\partial y}{\partial t})$

$P (x, t) = \{\begin{array}{cc} - F (0) (0) = 0 & if (x - vt) < - L \\ - F (\frac{h}{L}) (- \frac{h v}{L}) = F v {(\frac{h}{L})}^{2} & for - L < (x - vt) < 0 \\ - F (- \frac{h}{L}) (\frac{h v}{L}) = F v {(\frac{h}{L})}^{2} & for 0 < (x - vt) < L \\ - F (0) (0) = 0 & for (x - vt) > L \end{array}$