What does the wave reflected from the constant-level reservoir do to the pressure?

It returns the pressure to H0 as a negative wave

It makes the pressure H0 - dH immediately

It leaves the pressure unchanged

2072 Magh · 2 marks

Pressure-time diagram at 2/3 L from valve

Short answer

At 2/3 L from the valve, pressure remains static H0 for 0–2L/3c, then rises to H0+ΔH until 4L/3c. It returns to H0 until 8L/3c, drops to H0−ΔH from 8L/3c to 10L/3c, then H0 again, repeating every 4L/c.

In the figure below, water flowing through a pipe from the reservoir is suddenly stopped by closing a valve at point B. Draw pressure-time diagram at the 2/3 L from valve of the pipe for one cycle of wave motion. [Figure: reservoir connected to a pipe of length L ending in a valve at B]

Case. Reservoir at the upstream end, valve at B (downstream end), pipe length $L$ , wave speed $c$ , valve closed suddenly at $t = 0$ . The point is $\tfrac{2}{3}L$ from the valve (so $\tfrac{1}{3}L$ from the reservoir). Let $\Delta H = cV_0/g$ and the static head be $H_0$ .

Wave timing at the point ( $d = 2L/3$ from the valve)

$t_1 = \dfrac{d}{c} = \dfrac{2L}{3c}$ : the positive wave from the valve arrives; pressure rises to $H_0 + \Delta H$
$t_2 = \dfrac{2L - d}{c} = \dfrac{4L}{3c}$ : the negative wave reflected from the reservoir arrives; pressure returns to $H_0$
$t_3 = \dfrac{2L + d}{c} = \dfrac{8L}{3c}$ : the negative wave reflected from the valve arrives; pressure falls to $H_0 - \Delta H$
$t_4 = \dfrac{4L - d}{c} = \dfrac{10L}{3c}$ : the positive wave reflected from the reservoir arrives; pressure returns to $H_0$
At $t = \dfrac{4L}{c}$ the cycle is complete and repeats.

Interval	Pressure at the point
$0$ to $\dfrac{2L}{3c}$	$H_0$
$\dfrac{2L}{3c}$ to $\dfrac{4L}{3c}$	$H_0 + \Delta H$
$\dfrac{4L}{3c}$ to $\dfrac{8L}{3c}$	$H_0$
$\dfrac{8L}{3c}$ to $\dfrac{10L}{3c}$	$H_0 - \Delta H$
$\dfrac{10L}{3c}$ to $\dfrac{4L}{c}$	$H_0$

Pressure-time diagram (one cycle, $t$ from 0 to $4L/c$ )

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(top = 1.0, bottom = -1.0; '-' is zero line)

(+1 = $H_0 + \Delta H$ , 0 = $H_0$ , $-1$ = $H_0 - \Delta H$ .) The positive pulse lasts $2L/3c$ and the negative pulse lasts $2L/3c$ , separated by a period of $4L/3c$ at the static pressure.

✓ Answer checked by Claude Opus 5.5 (high effort)

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