Waves

JEE Main level, numerical answer, medium

64 of 142

Question 64 of 142

In an experiment with sonometer when a mass of 180g180g is attached to the string, it vibrates with fundamental frequency of 30Hz30Hz. When a mass mm is attached, the string vibrates with fundamental frequency of 50Hz50Hz. The value of mm is ______ gg.

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Answer: 500

The formula to calculate the fundamental frequency in a stretched string is given by
f=12LTμ...(1)f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}...\left(1\right)
For the first mass, it can be written that
f1=12Lm1gμ...(2){f}_{1}=\dfrac{1}{2L}\sqrt{\dfrac{{m}_{1}g}{\mu }}...\left(2\right)
And, for the second mass, it can be written that
f2=12Lm2gμ...(3){f}_{2}=\dfrac{1}{2L}\sqrt{\dfrac{{m}_{2}g}{\mu }}...\left(3\right)
Divide equation (3) by equation (2) and simplify to obtain the required mass.
f2f1=12Lm2gμ12Lm1gμ=m2m1⇒m1m2=(f1f2)2⇒m2=m1(f2f1)2...(4)\begin{matrix}\dfrac{{f}_{2}}{{f}_{1}} & = & \dfrac{\dfrac{1}{2L}\sqrt{\dfrac{{m}_{2}g}{\mu }}}{\dfrac{1}{2L}\sqrt{\dfrac{{m}_{1}g}{\mu }}} \\ & = & \sqrt{\dfrac{{m}_{2}}{{m}_{1}}} \\ & \Rightarrow & \dfrac{{m}_{1}}{{m}_{2}}={\left(\dfrac{{f}_{1}}{{f}_{2}}\right)}^{2} \\ & \Rightarrow & {m}_{2}={m}_{1}{\left(\dfrac{{f}_{2}}{{f}_{1}}\right)}^{2}...\left(4\right)\end{matrix}
Substitute the values of the known parameters into equation (4) to calculate the required mass.
m2=180g×(50Hz30Hz)2=500g\begin{matrix}{m}_{2} & = & 180g\times {\left(\dfrac{50Hz}{30Hz}\right)}^{2} \\ & = & 500g\end{matrix}

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