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Question
Sound wave A has period 0.015 s, sound wave B has a period of 0.025 s. Which sound has greater frequency?
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Solution
Given: TA = 0.015 s, TB = 0.025 s
To find: greater frequency (n)
Formula: n = `1/"T"`
Calculation: From formula,
`"n"_"A" = 1/"T"_"A"`
`= 1/0.015`
`= 1/(1.5 xx 10^-2)`
∴ nA = 66.67 Hz ...(using reciprocal table)
`"n"_"B" = 1/"T"_"B"`
`= 1/0.025`
`= 1/(2.5 xx 10^-2)`
∴ nB = 40 Hz ...(using reciprocal table)
∴ nA > nB
Sound wave A has a greater frequency than sound B.
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