Question

A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When

Answer 1

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

$\mathbf{f = \dfrac{v}{2L}}$

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

$\mathbf{254 - \dfrac{v}{2L} }$

These combinations of frequency produce 4 beats per sound.

i.e.

$\mathbf{254 - \dfrac{v}{2L} =4}$

$\mathbf{ \dfrac{v}{2L} = 254-4 }$

$\mathbf{ \dfrac{v}{2L} = 250 ----(1)}$

When it is altered, the beats first diminish and increase again by 4.

i.e.

$\mathbf{ \dfrac{v'}{2L} = 254+4 }$

$\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }$

If we equate both equations (1) and (2) together, we have:

$\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}$

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

• when the temperature of the pipe  = unknown ???
• the temperature of the open orang pipe = 15

∴

$\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}$

By squaring both sides, we have:

$\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}$

$\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}$

$\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}$

$\implies \mathbf{273 +T =306.726912 }$

T = 306.726912 - 273

T ≅ 33.73 ° C

∴

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

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