Question

A string under a tension of $ 129.6 $ N produces 10 beats/sec when it is vibrated along with a tuning fork. When the tension in the string is increased to 160N, it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork.
(A) 100Hz
(B) 50Hz
(C) 150Hz
(D) 200Hz

Answer 1

Hint: In a physics, tension is described as the pulling force transmitted axially by the means of a string, a cable, a chain, or similar one dimensional continuous object, or by each end of a rod. Tension could be opposite of compression. When the tension is increased to 160N, there is a resonance in frequency of the vibrating string and the tuning fork.
Resonance occurs when a system is able to store and easily transfer energy between different storage modes, such as kinetic energy or potential energy as you would find with a simple pendulum.
The frequency of the string is given by:
 $\Rightarrow v=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}} $
Where m is the mass of the string,
l is the length of the string,
T is the tension on the string applied.

Complete step by step solution
Let $ {{v}_{0}} $ be the resonant frequency of the vibrating string and the tuning fork.
In first condition;
No. of beats; b= 10 beats/sec
 $ \begin{align}
 &\Rightarrow {{v}_{1}}=\dfrac{1}{2l}\sqrt{\dfrac{{{T}_{1}}}{m}} \\
 &\Rightarrow {{v}_{0}}-10=\dfrac{1}{2l}\sqrt{\dfrac{{{T}_{1}}}{m}}\text{ }...................\text{ (1)} \\
\end{align} $
In second condition;
As there are no beats so there is only resonant frequency. So,
 $\Rightarrow {{v}_{0}}=\dfrac{1}{2l}\sqrt{\dfrac{{{T}_{2}}}{m}}\text{ }.........................\text{ (2)} $
Dividing equation (1) by equation (2)
 $\Rightarrow \dfrac{{{v}_{0}}-10}{{{v}_{0}}}=\dfrac{\dfrac{1}{2l}\sqrt{\dfrac{{{T}_{1}}}{m}}\text{ }}{\dfrac{1}{2l}\sqrt{\dfrac{{{T}_{2}}}{m}}} $
 $\Rightarrow \dfrac{{{v}_{0}}-10}{{{v}_{0}}}=\sqrt{\dfrac{{{T}_{1}}}{{{T}_{2}}}} $
 $\Rightarrow \dfrac{{{v}_{0}}-10}{{{v}_{0}}}=\sqrt{\dfrac{129.6}{160}} $
 $\Rightarrow \dfrac{{{v}_{0}}-10}{{{v}_{0}}}=\sqrt{\dfrac{81}{100}} $
 $\Rightarrow \dfrac{{{v}_{0}}-10}{{{v}_{0}}}=\dfrac{9}{10} $
 $\Rightarrow 10\left( {{v}_{0}}-10 \right)=9{{v}_{0}} $
 $\Rightarrow 10{{v}_{0}}-100=9{{v}_{0}} $
 $\Rightarrow 10{{v}_{0}}-9{{v}_{0}}=100 $
 $\Rightarrow {{v}_{0}}=100Hz $
Therefore, the fundamental frequency is 100Hz i.e. option (A).

Note
The value of this fundamental equation can also be found by forming the quadratic equation and solving it for the value of $ {{v}_{0}} $ . The fundamental frequency provides the sound with its strongest audible pitch reference - it is the predominant frequency in any complex waveform. Remember that there are four properties of string that affect its frequency are length, diameter, tension, and density. Tuning forks are available in a wide range of frequencies (64 Hz to 4096 Hz);128 Hz is commonly used for screening. We can increase the frequency of the tuning fork. The pitch that a particular tuning fork generates depends on the length of the prongs. Shorter prongs produce higher pitch(frequency) sound than longer prongs. This tuning fork is also used in the laboratory.