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A set of 24 tunning forks is arranged so that each gives 4 beats per second with the previous one and the last sounds the octave of first. Find frequency of 1st and last tunning forks. |
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Answer» Let frequency of Ist tunning fork = x frequency of IInd tunning fork = x + 4 frequency of IIIrd tunning fork = x + 2 (4) frequency of IVth tunning fork = x + 3 (4) ∴ Let frequency of 24th tunning fork = x + 23 (4) octave means, (twice in freq.) ∴ freq. of 24th = 2 × freq. of Ist = 2x ∴ 2x = x + 23 (4) ⇒ x = 92 freq. of 24th = 2 × 92 = 184 H3. |
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