
For a stretched string of given length, the tension $T$ is plotted on the X-axis and the frequency $f$ on the Y-axis. The graph has the shape of a
A) Rectangular Hyperbola
B) Straight through the origin
C) Parabola
D) Straight line not through the origin
Answer
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Hint: Whenever a string is stretched under tension, transverse waves can travel on the string. Use the formula of the speed of waves on a string and use the relation between frequency and speed to determine the shape of the graph.
Formula used:
$\Rightarrow v = \sqrt {\dfrac{T}{\mu }} $ where $v$ is the velocity of waves on a string, $T$ is the tension on the string, and $\mu $ is the linear mass density of the string.
$\Rightarrow v = L \times f$where $L$ is the length of the string and $f$ is the frequency of the waves traveling on the string
Complete step by step solution:
A wave can travel on a string under tension with a speed:
$\Rightarrow v = \sqrt {\dfrac{T}{\mu }} $
Since we’ve been asked to find the relation between frequency and Tension, we will use the relation between velocity and frequency $v = L \times f$ where$L$ is the length of the string and$f$ is the frequency of the waves traveling on the string.
Substituting the value of $v = \sqrt {\dfrac{T}{\mu }} $ in $v = L \times f$, we get:
$\Rightarrow \sqrt {\dfrac{T}{\mu }} = L \times f$
On squaring both sides, we get
$\Rightarrow \dfrac{T}{\mu } = {L^2} \times {f^2}$
Dividing both sides by ${L^2}$ , we get:
$\Rightarrow {f^2} = \dfrac{T}{{\mu {L^2}}}$
This equation can be written in the form of
$\Rightarrow {f^2} = 4aT$ , where $a = \dfrac{1}{{4\mu {L^2}}}$ is a constant dependent on string mass and length.
This equation has the form of
$\Rightarrow {y^2} = 4ax$ which is the equation of a parabola.
Hence If we plotted the tension $T$on the X-axis and the frequency $f$on the Y –axis, we will have the shape of a parabola which corresponds to option (C).
Note:
The speed of waves on a string is proportional to the tension on the string. So for higher tension on the string, the frequency of the waves travelling on the sting will be higher. This property is often used in musical instruments like the guitar to generate different sounds with different frequencies by tightening the string to increase tension on it (pitch).
Formula used:
$\Rightarrow v = \sqrt {\dfrac{T}{\mu }} $ where $v$ is the velocity of waves on a string, $T$ is the tension on the string, and $\mu $ is the linear mass density of the string.
$\Rightarrow v = L \times f$where $L$ is the length of the string and $f$ is the frequency of the waves traveling on the string
Complete step by step solution:
A wave can travel on a string under tension with a speed:
$\Rightarrow v = \sqrt {\dfrac{T}{\mu }} $
Since we’ve been asked to find the relation between frequency and Tension, we will use the relation between velocity and frequency $v = L \times f$ where$L$ is the length of the string and$f$ is the frequency of the waves traveling on the string.
Substituting the value of $v = \sqrt {\dfrac{T}{\mu }} $ in $v = L \times f$, we get:
$\Rightarrow \sqrt {\dfrac{T}{\mu }} = L \times f$
On squaring both sides, we get
$\Rightarrow \dfrac{T}{\mu } = {L^2} \times {f^2}$
Dividing both sides by ${L^2}$ , we get:
$\Rightarrow {f^2} = \dfrac{T}{{\mu {L^2}}}$
This equation can be written in the form of
$\Rightarrow {f^2} = 4aT$ , where $a = \dfrac{1}{{4\mu {L^2}}}$ is a constant dependent on string mass and length.
This equation has the form of
$\Rightarrow {y^2} = 4ax$ which is the equation of a parabola.
Hence If we plotted the tension $T$on the X-axis and the frequency $f$on the Y –axis, we will have the shape of a parabola which corresponds to option (C).
Note:
The speed of waves on a string is proportional to the tension on the string. So for higher tension on the string, the frequency of the waves travelling on the sting will be higher. This property is often used in musical instruments like the guitar to generate different sounds with different frequencies by tightening the string to increase tension on it (pitch).
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