Question

A sinewave has a period (duration of one cycle) of 6.5ms. What is its corresponding frequency expressed to 3 significant figures? If the frequency of this sinewave is increased by a factor of 3, calculate the new period value, expressing your answer again to 3 significant figures. Clearly show all your working.

Answer 1

Answer:

(A) Corresponding frequency = 154 Hz (in 3 significant figures)

(B) New Period = 0.00217 s (in 3 significant figures)

Explanation:

The given information is as follows:

The period of a sinewave =  $T = 6.5\ ms = 6.5\ * \ 10^{-3}\ s = 0.0065s$

Part-A

Corresponding frequency $f$ is required.

Since,

$frequency = \frac{1}{Period} \\f = \frac{1}{T}$

Plug the value of the period, $T$, of a given sine wave in the above equation:

$f = \frac{1}{0.0065} \\f= 153.846\ Hz$

The aforementioned frequency should be expressed to 3 significant figures (which is the requirement of the question). So, take the first three digits of the answer (since they are significant), but as you can see after the decimal point, the digit is greater than 5, round the third digit off to 4. Hence, the corresponding frequency would be 154 Hz (expressed in 3 significant figures).

Part-B

The frequency of the sinewave is increased by a factor of 3. It means that you need to multiply the frequency found in the Part-A by 3.

Hence, the new frequency will be as follows:

$f_{new} = 3*f_{old} \\f_{new} = 3*153.846\\f_{new}= 461.538\ Hz$

Note that for finding the new frequency, the old, unrounded off frequency is used.

Since,

$New\ frequency = \frac{1}{New Period} \\f_{new} = \frac{1}{T_{new}}\\T_{new} = \frac{1}{f_{new}}$

Plug the value of the new frequency in the above equation, and evaluate for the new period:

$T_{new} = \frac{1}{461.538}}\\T_{new} = 0.00216667\ s$

Again, round it off to three significant figures. Zeroes before 2 do not matter when we count for the significant figures (a significant figure rule). Also make sure to round off the third digit from 2 (since the digit next to that digit is greater than 5).

Hence, the new period value will be 0.00217s (expressed in 3 significant figures).