A tuning fork gives 4beats with 50cm length of a Sonometer wire. If the length of the wire is shortened by 1cm, then the number of beats is still the same. Then, the frequency of the fork is:
(A) 396Hz
(B) 400Hz
(c) 404Hz
(D) 384Hz
Answer
Verified
423.6k+ views
Hint: We shall use the basic property of a tuning fork, that is, the frequency of the tuning fork is inversely proportional to the length of the string. Since, the number of beats is the same before and after making a change in the length of the string, it means the change in frequency for beats to be observed is the same.
Complete answer:
Let the frequency of the tuning fork be $f$and let the frequency initially for 4 beats be ${{f}_{1}}$ .
Then, it has been given to us in the question that:
$\Rightarrow f-{{f}_{1}}=4$
$\Rightarrow {{f}_{1}}=f-4$ [Let this expression be equation number (1)]
Let the frequency after cutting the wire by 1cm be ${{f}_{2}}$ .
We know that the frequency of the tuning fork is inversely proportional to the length of the wire, that is:
$\Rightarrow f\propto \dfrac{1}{l}$
$\Rightarrow fl=constant$
Now, in the second case the length has been decreased, this implies that the frequency has increased.
As it's been given in the problem that, the number of beats is the same as 4. Therefore:
$\Rightarrow {{f}_{2}}-f=4$
$\Rightarrow {{f}_{2}}=f+4$ [Let this expression be equation number (2)]
From equation number (1) and (2), we have:
$\Rightarrow {{f}_{1}}{{l}_{1}}={{f}_{2}}{{l}_{2}}$
Where, it has been given that:
$\begin{align}
& \Rightarrow {{l}_{1}}=50cm \\
& \Rightarrow {{l}_{2}}=49cm \\
\end{align}$
Putting the values of all these terms in the above equation, we get the frequency of tuning fork as:
$\begin{align}
& \Rightarrow (f-4)50=(f+4)49 \\
& \Rightarrow 50f-49f=200+196 \\
& \Rightarrow f=396Hz \\
\end{align}$
Hence, the frequency of the tuning fork comes out to be 396Hz.
Hence, option (A) is the correct option.
Note:
Our conclusion of the fact that the frequency afterwards was increased was based on a simple property that, length of wire is inversely proportional to frequency. These are some key properties regarding tuning forks. If we had not known this property and assumed the number of beats is the same so the frequency must be the same, then we might have ended up solving an equation which had no result.
Complete answer:
Let the frequency of the tuning fork be $f$and let the frequency initially for 4 beats be ${{f}_{1}}$ .
Then, it has been given to us in the question that:
$\Rightarrow f-{{f}_{1}}=4$
$\Rightarrow {{f}_{1}}=f-4$ [Let this expression be equation number (1)]
Let the frequency after cutting the wire by 1cm be ${{f}_{2}}$ .
We know that the frequency of the tuning fork is inversely proportional to the length of the wire, that is:
$\Rightarrow f\propto \dfrac{1}{l}$
$\Rightarrow fl=constant$
Now, in the second case the length has been decreased, this implies that the frequency has increased.
As it's been given in the problem that, the number of beats is the same as 4. Therefore:
$\Rightarrow {{f}_{2}}-f=4$
$\Rightarrow {{f}_{2}}=f+4$ [Let this expression be equation number (2)]
From equation number (1) and (2), we have:
$\Rightarrow {{f}_{1}}{{l}_{1}}={{f}_{2}}{{l}_{2}}$
Where, it has been given that:
$\begin{align}
& \Rightarrow {{l}_{1}}=50cm \\
& \Rightarrow {{l}_{2}}=49cm \\
\end{align}$
Putting the values of all these terms in the above equation, we get the frequency of tuning fork as:
$\begin{align}
& \Rightarrow (f-4)50=(f+4)49 \\
& \Rightarrow 50f-49f=200+196 \\
& \Rightarrow f=396Hz \\
\end{align}$
Hence, the frequency of the tuning fork comes out to be 396Hz.
Hence, option (A) is the correct option.
Note:
Our conclusion of the fact that the frequency afterwards was increased was based on a simple property that, length of wire is inversely proportional to frequency. These are some key properties regarding tuning forks. If we had not known this property and assumed the number of beats is the same so the frequency must be the same, then we might have ended up solving an equation which had no result.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
The reservoir of dam is called Govind Sagar A Jayakwadi class 11 social science CBSE
10 examples of friction in our daily life
What problem did Carter face when he reached the mummy class 11 english CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE