
Audible frequency range of a human ear is $20\,Hz$ to $20000\,Hz$ . Express it in terms of time period?
Answer
585.3k+ views
Hint:You can approach the solution to this question by trying to recall the relation between time period and frequency. Frequency is the occurrence of an event in a unit second while time period is the time taken to complete one cycle of a repeating event. Find the mathematical relation between the two and you’ll reach the answer easily.
Complete step by step answer:
We will try to solve the question exactly like we explained in the hint section of the solution to this question. Firstly, we will find some relation between the time period of an event and frequency of an event. Once we do that, we will express the relation mathematically and use that to reach at the answer to the question mentioned above.
First, let us define frequency and time period:
Frequency: It is the measure of number of occurrences of a cycle of a repeating even in a unit second, i.e. $1\,s$
Time period: Time period is the time taken by a cycle of a repeating event to complete fully.
We can see that there is definitely some relation between frequency and time period, both as somehow complementary of each other, and we can write the relation mathematically as:
$T = \dfrac{1}{f}$
Where, $T$ is the time period and $f$ is the frequency.
We have been given the range of frequency of sound that humans can hear. The question has asked us to convert this range in terms of time period, which we can easily do using the relation between time period and frequency that we just found out.
Lower limit of time period, ${T_l} = \dfrac{1}{{{f_u}}}$ ( ${f_u}$ is the upper limit of frequency)
Substituting the value given to us, we get:
$
{T_l} = \dfrac{1}{{20000}}\,s \\
{T_l} = 5 \times {10^{ - 5}}\,s \\
$
Now, we need to find the upper limit of time period:
${T_u} = \dfrac{1}{{{f_l}}}$ ( ${f_l}$ is the lower limit of frequency)
Substituting the value given to us in the question, we get:
$
{T_u} = \dfrac{1}{{20}}\,s \\
{T_u} = 0.05\,s \\
$
So, we can say that the range of sound that humans can hear in terms of time period is $5 \times {10^{ - 5}}\,s$ to $0.05\,s$ .
Note:Point that should be noted is that the upper limit of frequency doesn’t represent the upper limit of time period as well, since both are reciprocal of each other, the upper limit of frequency leads us to lower limit of time period and vice versa.
Complete step by step answer:
We will try to solve the question exactly like we explained in the hint section of the solution to this question. Firstly, we will find some relation between the time period of an event and frequency of an event. Once we do that, we will express the relation mathematically and use that to reach at the answer to the question mentioned above.
First, let us define frequency and time period:
Frequency: It is the measure of number of occurrences of a cycle of a repeating even in a unit second, i.e. $1\,s$
Time period: Time period is the time taken by a cycle of a repeating event to complete fully.
We can see that there is definitely some relation between frequency and time period, both as somehow complementary of each other, and we can write the relation mathematically as:
$T = \dfrac{1}{f}$
Where, $T$ is the time period and $f$ is the frequency.
We have been given the range of frequency of sound that humans can hear. The question has asked us to convert this range in terms of time period, which we can easily do using the relation between time period and frequency that we just found out.
Lower limit of time period, ${T_l} = \dfrac{1}{{{f_u}}}$ ( ${f_u}$ is the upper limit of frequency)
Substituting the value given to us, we get:
$
{T_l} = \dfrac{1}{{20000}}\,s \\
{T_l} = 5 \times {10^{ - 5}}\,s \\
$
Now, we need to find the upper limit of time period:
${T_u} = \dfrac{1}{{{f_l}}}$ ( ${f_l}$ is the lower limit of frequency)
Substituting the value given to us in the question, we get:
$
{T_u} = \dfrac{1}{{20}}\,s \\
{T_u} = 0.05\,s \\
$
So, we can say that the range of sound that humans can hear in terms of time period is $5 \times {10^{ - 5}}\,s$ to $0.05\,s$ .
Note:Point that should be noted is that the upper limit of frequency doesn’t represent the upper limit of time period as well, since both are reciprocal of each other, the upper limit of frequency leads us to lower limit of time period and vice versa.
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