Question

The water level in a vertical glass tube$\mathbf{\text{1.00m}}$long can be adjusted to any position in the tube. A tuning fork vibrating at$\mathbf{\text{686Hz}}$is held just over the open top end of the tube, to set up a standing wave of sound in the air-filled top portion of the tube. (That air-filled top portion acts as a tube with one closed and the other end open)(a) For how many different positions of the water level will sound from the fork set up resonance in the tube’s air-filled portion, which acts as a pipe with one end closed (by the water) and the other end open? What are the (b) least (c) second least water heights in the tube for resonance to occur?

Short answer

a. The different positions of water are$0.125\mathrm{m},0.375\mathrm{m},0.625\mathrm{m},0.875\mathrm{m}$

b. The least water heights in the tube for resonance to occur is $0.125\mathrm{m}$.

c. The second least water heights in the tube for resonance to occur is$0.375\mathrm{m}$

Step-by-step solution

Given

• Length of glass tube ,$\mathrm{L}=1.\text{00m}$
• Frequency of the tuning fork ,$\mathrm{f}=68\text{6Hz}$

Determining the concept

Apply the formula for the harmonics (frequency) for a closed organ pipe to find the different positions of water levels.

Formula is as follow:

$\mathrm{f}=\left(\mathrm{n}-\frac{1}{2}\right)\left(\frac{\mathrm{v}}{2\mathrm{X}}\right)$

Here, $\mathrm{f}$is frequency,$\mathrm{x}$ is the position and$\mathrm{v}$is the velocity.

(a) Determine the positions of the water

Frequency (resonance) of closed pipe is defined as,

$\mathrm{f}=\left(\mathrm{n}-\frac{1}{2}\right)\left(\frac{\mathrm{v}}{2\mathrm{X}}\right)$

Here, $\mathrm{x}$is the distance from the top of the tube,

$68\text{6Hz}=\left(\mathrm{n}-\frac{1}{2}\right)\left(\frac{343\mathrm{m}/\mathrm{s}}{2\mathrm{X}}\right)$

$\mathrm{X}=\left(\mathrm{n}-\frac{1}{2}\right)\left(\frac{1}{4}\right)$

When,$\mathrm{n}=1$,

$\begin{array}{c}\mathrm{X}=\left(1-\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.12\text{5m}\end{array}$

When, $\mathrm{n}=2$

$\begin{array}{c}\mathrm{X}=\left(2-\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.375\mathrm{m}\end{array}$

When,$\mathrm{n}=3$

$\begin{array}{c}\mathrm{X}=\left(3-\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.62\text{5m}\end{array}$

When,$\mathrm{n}=4$

$\begin{array}{c}\mathrm{X}=\left(4-\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.875\text{\hspace{0.17em}m}\end{array}$

Hence, these are the different positions of water.

(b) Determine the least water heights in the tube for resonance to occur

For the height of water, measure the height from the bottom of the tube,

$\begin{array}{c}{\mathrm{h}}_1=1\text{\hspace{0.17em}m}-0.875\text{m}\\ =0.\text{125m}\end{array}$

Hence, the least water heights in the tube for resonance to occur is $0.\text{125m}$.

(c) Determining the second least water heights in the tube for resonance to occur

Second least height

$\begin{array}{c}{\mathrm{h}}_2=1\text{\hspace{0.17em}m}-0.62\text{5m}\\ =0.37\text{5m}\end{array}$

Hence, the second least water heights in the tube for resonance to occur is $0.37\text{5m}$.