Answer
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Hint Frequency is the number of cycles per second meant as Hertz. The period is seconds per cycle. Converse since, supposing that the frequency is high, at that point the period is low. So by using the above statement we will be able to find it.
Formula used:
Time period,
$T = \dfrac{1}{f}$
Here,
$T$, will be the time period
$f$, will be the frequency.
Complete Step By Step Solution Frequency and period are contrary related amounts. So, the time span between progressive compressions is equivalent to the time-frame of the wave. Furthermore, this time span is proportional to the recurrence of the wave.
Therefore, by using the formula for the time period
We get
$T = \dfrac{1}{f}$
Now substitute the values, we get
$ \Rightarrow T = \dfrac{1}{{500}}s$
Now on solving the above equation, we get
$ \Rightarrow T = 0.002{\text{ s}}$
Therefore, in milliseconds it will be
$ \Rightarrow T = 2{\text{ ms}}$
The above time period will be required for the source having successive compression.
Hence the option $\left( c \right)$ will be correct.
Additional information At the point when a function happens consistently, at that point, we state that the function is occasional and allude to the ideal opportunity for the function to rehash itself as the period. Period - as expected - is estimated in a flash, hours, days, or years. For example, the time of the pivot of Earth on its hub is twenty-four hours.
Note There is the relationship that the time period is the average distance between bodies or particles or photons divided by their average velocity. The frequency is the reciprocal of the period. The average is measured over some time if it varies, or it is necessary to do so to measure a minimum quantity.
Formula used:
Time period,
$T = \dfrac{1}{f}$
Here,
$T$, will be the time period
$f$, will be the frequency.
Complete Step By Step Solution Frequency and period are contrary related amounts. So, the time span between progressive compressions is equivalent to the time-frame of the wave. Furthermore, this time span is proportional to the recurrence of the wave.
Therefore, by using the formula for the time period
We get
$T = \dfrac{1}{f}$
Now substitute the values, we get
$ \Rightarrow T = \dfrac{1}{{500}}s$
Now on solving the above equation, we get
$ \Rightarrow T = 0.002{\text{ s}}$
Therefore, in milliseconds it will be
$ \Rightarrow T = 2{\text{ ms}}$
The above time period will be required for the source having successive compression.
Hence the option $\left( c \right)$ will be correct.
Additional information At the point when a function happens consistently, at that point, we state that the function is occasional and allude to the ideal opportunity for the function to rehash itself as the period. Period - as expected - is estimated in a flash, hours, days, or years. For example, the time of the pivot of Earth on its hub is twenty-four hours.
Note There is the relationship that the time period is the average distance between bodies or particles or photons divided by their average velocity. The frequency is the reciprocal of the period. The average is measured over some time if it varies, or it is necessary to do so to measure a minimum quantity.
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