This 2007 IIT JEE physics subjective practice problem requires the student to find the end correction, or difference between the frequency of a tuning fork and the corresponding sound waves inside of a tube, given the contextual information. This type of problem has an answer between 0 and 9999, with six points awarded for a correct answer. No points are subtracted for an incorrect response, so all problems in the subjective category may be attempted without penalty.
In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode with a tuning fork is 0.30 m. When this length is changed to 1.00 m, the same tuning fork resonates with the first overtone. What is the end correction (in cm) for the column?
In the resonance column method, the pipe has one end closed and one end open. Let Δl be the end correction. Remember that the first resonance is, theoretically, one quarter of a wavelength, with the distance between resonances at half of a wavelength.
Thus, for the fundamental mode: 0.30 m + Δl = λ/4
For the first overtone: 1.00 m + Δl = 3λ/4
⇒ 3(0.30 m + Δl) = 1.00 m + Δl
⇒ Δl = 0.05 m = 5 cm.
Bio: Alexis Bonari is a freelance writer and blog junkie. She often can be found blogging about general education issues as well as information on college scholarships. In her spare time, she enjoys square-foot gardening, swimming, and avoiding her laptop.
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