
A person can hear frequencies only below $10kHz$. A steel piano wire $50cm$ long of mass $5g$ is stretched with a tension of $400N$. The number of the highest overtone of the sound produced by this piano wire that the person can hear is
$
A.{\text{ 48}} \\
B.{\text{ 50}} \\
{\text{C}}{\text{. 49}} \\
{\text{D}}{\text{. 51}} \\
$
Answer
584.7k+ views
Hint: Humans can detect and hear sounds in the frequency range from about $20Hz{\text{ to 20kHz}}$. Actually, human infants can hear frequencies slightly higher than $20kHz$ but tend to lose the high frequency sensitivity as they get mature, the upper limit in the average adult is most often closer to the $15 - 17kHz$. Here first convert given terms in the same unit system and solve using the standard formula.
Formula used:
-$\mu = \dfrac{M}{L}$
Where, $\mu $ is the mass per unit Length of the string.
-\[v = \dfrac{n}{{2l}}\sqrt {\dfrac{T}{\mu }} \]
Here, $v$ is the frequency, $T$ is the tension and $n$ is the number of overtones.
Complete step by step answer:
$\mu = \dfrac{{5 \times {{10}^{ - 3}}}}{{0.5}}$
Simplify the above equation using the basic mathematical property
$
\mu = \dfrac{{5 \times {{10}^{ - 3}} \times 10}}{5} \\
\mu = 0.01 \\
$
$T = 400N$
Frequency, \[v = \dfrac{n}{{2l}}\sqrt {\dfrac{T}{\mu }} \]
Place the known values –
\[v = \dfrac{n}{{2 \times 0.5}}\sqrt {\dfrac{{400}}{{0.01}}} \]
Simplify the left hand side of the equation –
\[
v = \dfrac{n}{1}\sqrt {\dfrac{{400 \times 100}}{1}} \\
v = n \times 200 \\
v = 200n \\
\]
Also, a person can hear frequencies below$10kHz$only.
Place value in the above equation –
$
10000 > 200n \\
\therefore \dfrac{{10000}}{{200}} > n \\
\therefore 50 > n \\
$
Therefore, the number of the highest overtones of the sound produced by this piano wire that the person can hear is $49$.
Hence, from the given multiple choices – the option C is the correct answer.
Note: Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.
Formula used:
-$\mu = \dfrac{M}{L}$
Where, $\mu $ is the mass per unit Length of the string.
-\[v = \dfrac{n}{{2l}}\sqrt {\dfrac{T}{\mu }} \]
Here, $v$ is the frequency, $T$ is the tension and $n$ is the number of overtones.
Complete step by step answer:
$\mu = \dfrac{{5 \times {{10}^{ - 3}}}}{{0.5}}$
Simplify the above equation using the basic mathematical property
$
\mu = \dfrac{{5 \times {{10}^{ - 3}} \times 10}}{5} \\
\mu = 0.01 \\
$
$T = 400N$
Frequency, \[v = \dfrac{n}{{2l}}\sqrt {\dfrac{T}{\mu }} \]
Place the known values –
\[v = \dfrac{n}{{2 \times 0.5}}\sqrt {\dfrac{{400}}{{0.01}}} \]
Simplify the left hand side of the equation –
\[
v = \dfrac{n}{1}\sqrt {\dfrac{{400 \times 100}}{1}} \\
v = n \times 200 \\
v = 200n \\
\]
Also, a person can hear frequencies below$10kHz$only.
Place value in the above equation –
$
10000 > 200n \\
\therefore \dfrac{{10000}}{{200}} > n \\
\therefore 50 > n \\
$
Therefore, the number of the highest overtones of the sound produced by this piano wire that the person can hear is $49$.
Hence, from the given multiple choices – the option C is the correct answer.
Note: Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.
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