JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 13)
A tuning fork vibrates with frequency $$256$$ $$Hz$$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe ? (Speed of sound in air is $$340\,m{s^{ - 1}}$$)
$$220$$ $$cm$$
$$190$$ $$cm$$
$$180$$ $$cm$$
$$200$$ $$cm$$
Explanation
The tuning fork vibrates with frequency 256 Hz and give one beat per second So, the organ pipe will have frequency (256 $$ \pm $$ 1) Hr.
For open organ pipe,
Frequency n = $${{N\upsilon } \over {2\ell }}$$
Here n = 255 Hz
N = 3
$$\upsilon $$ = 340 m/s
$$\therefore\,\,\,\,$$ 255 = $${{3 \times 340} \over {2 \times \ell }}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $$\ell $$ = $${{3 \times 340} \over {2 \times 255}} = 2\,m$$
$$\therefore\,\,\,\,$$ $$\ell $$ = 2m or 200 cm
For open organ pipe,
Frequency n = $${{N\upsilon } \over {2\ell }}$$
Here n = 255 Hz
N = 3
$$\upsilon $$ = 340 m/s
$$\therefore\,\,\,\,$$ 255 = $${{3 \times 340} \over {2 \times \ell }}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $$\ell $$ = $${{3 \times 340} \over {2 \times 255}} = 2\,m$$
$$\therefore\,\,\,\,$$ $$\ell $$ = 2m or 200 cm
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