Answer
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Hint: Tension in the string is defined as the product of the velocity of the string to the root of the ratio of the strings mass and its length. Put the above relation in the formula for frequency which in conjunction with the velocity of the string.
Complete step by step solution:
We know that energy is given as:
$E = h \vee $;
Here,
E = Energy;
h = Planck’s Constant;
$ \vee $= Frequency;
$ \vee = \dfrac{c}{\lambda }$;
Here:
c = Speed of light;
$\lambda $= Wavelength;
\[E = \dfrac{{hc}}{\lambda }\] ;
Now, Energy is also related with Planck’s constant as:
\[E = hf\];
Equate the above two equations together:
\[hf = \dfrac{{hc}}{\lambda }\];
\[ \Rightarrow f = \dfrac{c}{\lambda }\];
Here, c is the speed of light but in case of the stretched string the speed will be the velocity of the wave that is travelling through the string. So:
\[\Rightarrow f = \dfrac{{{v_{{\text{wave on string}}}}}}{\lambda }\];
Now, in the fundamental mode the wavelength is twice the length of the string:
$\Rightarrow f = \dfrac{{{v_{{\text{wave on string}}}}}}{{2L}}$;
Now, the relation of wave velocity and the tension of the string is given by:
$\Rightarrow {v_{{\text{wave on string}}}} = \sqrt {\dfrac{T}{{m/L}}} $;
Put the above relation in the formula for frequency:
$\Rightarrow f = \dfrac{{\sqrt {\dfrac{T}{{m/L}}} }}{{2L}}$;
Now, the above is the original frequency of the string in relation with Tension in the string, if we quadruple the tension in the string to four times then the change in the frequency will be:
$\Rightarrow {f_n} = \dfrac{{\sqrt {\dfrac{{4T}}{{m/L}}} }}{{2L}}$;
$\Rightarrow {f_n} = 2 \times \dfrac{{\sqrt {\dfrac{T}{{m/L}}} }}{{2L}}$;
Now, we have been given that the fundamental frequency is $f = \dfrac{{\sqrt {\dfrac{T}{{m/L}}} }}{{2L}}$:
$ \Rightarrow {f_n} = 2 \times f$;
Therefore, the change in the fundamental frequency of the transverse wave that can be created in the stretched string is that we can double the frequency (2f).
Note: Here, we need to derive the relation between the frequency and the velocity of the wave and then apply the relation of velocity and tension in the string and apply this relation in the frequency formula, after that apply the given conditions and compare the two frequencies.
Complete step by step solution:
We know that energy is given as:
$E = h \vee $;
Here,
E = Energy;
h = Planck’s Constant;
$ \vee $= Frequency;
$ \vee = \dfrac{c}{\lambda }$;
Here:
c = Speed of light;
$\lambda $= Wavelength;
\[E = \dfrac{{hc}}{\lambda }\] ;
Now, Energy is also related with Planck’s constant as:
\[E = hf\];
Equate the above two equations together:
\[hf = \dfrac{{hc}}{\lambda }\];
\[ \Rightarrow f = \dfrac{c}{\lambda }\];
Here, c is the speed of light but in case of the stretched string the speed will be the velocity of the wave that is travelling through the string. So:
\[\Rightarrow f = \dfrac{{{v_{{\text{wave on string}}}}}}{\lambda }\];
Now, in the fundamental mode the wavelength is twice the length of the string:
$\Rightarrow f = \dfrac{{{v_{{\text{wave on string}}}}}}{{2L}}$;
Now, the relation of wave velocity and the tension of the string is given by:
$\Rightarrow {v_{{\text{wave on string}}}} = \sqrt {\dfrac{T}{{m/L}}} $;
Put the above relation in the formula for frequency:
$\Rightarrow f = \dfrac{{\sqrt {\dfrac{T}{{m/L}}} }}{{2L}}$;
Now, the above is the original frequency of the string in relation with Tension in the string, if we quadruple the tension in the string to four times then the change in the frequency will be:
$\Rightarrow {f_n} = \dfrac{{\sqrt {\dfrac{{4T}}{{m/L}}} }}{{2L}}$;
$\Rightarrow {f_n} = 2 \times \dfrac{{\sqrt {\dfrac{T}{{m/L}}} }}{{2L}}$;
Now, we have been given that the fundamental frequency is $f = \dfrac{{\sqrt {\dfrac{T}{{m/L}}} }}{{2L}}$:
$ \Rightarrow {f_n} = 2 \times f$;
Therefore, the change in the fundamental frequency of the transverse wave that can be created in the stretched string is that we can double the frequency (2f).
Note: Here, we need to derive the relation between the frequency and the velocity of the wave and then apply the relation of velocity and tension in the string and apply this relation in the frequency formula, after that apply the given conditions and compare the two frequencies.
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