Question

Your boss asks you compare two sound waves and determine which has the higher frequency. Wave A has a period of 4⁄100 second and Wave B has a period of 1⁄7 second. What would you tell your boss?
A. Wave B has the higher frequency. Its frequency is 7 Hz.
B. Wave A has the higher frequency. Its frequency is 25 Hz.
C. Wave B has the higher frequency. Its frequency is 42 Hz.
D. Wave A has the higher frequency. Its frequency is 4 Hz.

Answer 1

Answer

B. Wave A has the higher frequency. Its frequency is 25 Hz.

Explanation

Frequency is the number of oscillation/revolutions in a unit time. On the other hand, period is the time taken to complete one oscillation/revolution.

From the definition above we can conclude that, frequency(f) is the reciprocal of period(T). That is;

f = 1/T and

T = 1/f

Wave A

f = 1/(4/100)

= 100/4

= 25 Hz

Wave B

f = 1/(1/7)

= 7/1

= 7 Hz

So, wave A has the higher frequency. Its frequency is 25 Hz.

Answer 2

Answer: The correct answer is Option B.

Explanation:

Frequency is defined as the reciprocal of time period. It is measured in Hertz(Hz) or $Sec^{-1}$.

$\nu=\frac{1}{\text{Time Period}}$

For wave A:

Time period given is $\frac{4}{100}seconds$. So, frequency will be:

$\nu_A=\frac{1}{4/100}=\frac{100}{4}=25Hz$

For wave B:

Time period given is $\frac{1}{7}seconds$. So, frequency will be:

$\nu_B=\frac{1}{1/7}=\frac{7}{1}=7Hz$

From above, we can say that the frequency of wave A is more than the frequency of wave B.

Hence, the correct answer is Option B.