The frequency of a tuning fork is 512 Hz. It produces resonance in a closed organ pipe of length 16 cm (speed of sound 340 m/s). The pipe resonates in MCQ with Answer and Explanation
The frequency of a tuning fork is 512 Hz. It produces resonance in a closed organ pipe of length 16 cm (speed of sound 340 m/s). The pipe resonates in
A. First overtone
B. Second overtone
C. End correction needed
D. Fundamental mode
Answer: Option D
Solution (By JKSSB Mock Tests)
For closed pipe, fundamental f = v/(4L) = 340/(4×0.16) = 340/0.64 ≈ 531 Hz, close to 512 Hz, slight difference due to end correction. So fundamental. Approx match, SSC type.
Explanation:
Work W = F·d = Fd cosθ. cos0° = 1 gives maximum work. At 90°, cos90° = 0, work zero. For given magnitudes, maximum when force parallel to displacement. This is basic definition.
Explanation:
Doppler effect: observed frequency higher when approaching, lower when receding. Applications: radar speed guns, astronomical redshift/blueshift, echocardiography. Echo is reflection. Resonance is vibration matching. Diffraction is bending.
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