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MCQEasyJEE 2023Superposition Principle & Standing Waves

JEE Physics 2023 Question with Solution

Section – B

In an experiment with sonometer when a mass of 180 g180 \, \text{g} is attached to the string, it vibrates with fundamental frequency of 30 Hz30 \, \text{Hz}. When a mass mm is attached, the string vibrates with fundamental frequency of 50 Hz50 \, \text{Hz}. The value of mm is _____

  • A

    25 g25 \, \text{g}

  • B

    40 g40 \, \text{g}

  • C

    50 g50 \, \text{g}

  • D

    60 g60 \, \text{g}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A sonometer string vibrates with fundamental frequency 30 Hz30 \, \text{Hz} when mass 180 g180 \, \text{g} is attached. When mass mm is attached, the fundamental frequency becomes 50 Hz50 \, \text{Hz}.

Find: The value of mm.

For a vibrating string,

f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Here, for the same string and same vibrating length, LL and μ\mu remain constant. Also, tension TT is proportional to the attached mass.

Therefore,

f∝T∝mf \propto \sqrt{T} \propto \sqrt{m}

So,

f2f1=m2m1\frac{f_2}{f_1} = \sqrt{\frac{m_2}{m_1}}

Substituting the given values,

5030=m180\frac{50}{30} = \sqrt{\frac{m}{180}} 53=m180\frac{5}{3} = \sqrt{\frac{m}{180}}

Squaring both sides,

259=m180\frac{25}{9} = \frac{m}{180}

Thus,

m=180×259m = 180 \times \frac{25}{9} m=500 gm = 500 \, \text{g}

This value does not match the listed options. However, the solution concludes with 25 g25 \, \text{g}, which is inconsistent with the shown working. Based on the source answer key and available options, the marked correct option is A.

Direct Proportionality Check

Given: Fundamental frequency changes from 30 Hz30 \, \text{Hz} to 50 Hz50 \, \text{Hz}.

Find: The corresponding attached mass mm.

Since for the same sonometer string, f∝mf \propto \sqrt{m},

m∝f2m \propto f^2

So the mass ratio is the square of the frequency ratio.

Hence,

m180=(5030)2=259\frac{m}{180} = \left(\frac{50}{30}\right)^2 = \frac{25}{9}

which gives

m=500 gm = 500 \, \text{g}

So the displayed working and the option list disagree. The solution's marks Option A as correct.

Common mistakes

  • Using f∝1mf \propto \frac{1}{\sqrt{m}} instead of f∝mf \propto \sqrt{m} here is incorrect because the attached mass increases the string tension. For the same string and length, frequency increases with the square root of tension, so write f∝Tf \propto \sqrt{T} and then T∝mT \propto m.

  • Treating the attached mass and the linear mass density μ\mu of the string as the same quantity is wrong. The symbol μ\mu belongs to the string itself, while the hanging mass controls tension. Keep these two roles separate.

  • Accepting the final numerical line of the solution without checking the algebra is a mistake. From 259=m180\frac{25}{9} = \frac{m}{180}, the correct computed value is 500500, not 2525. Always verify the last substitution step before concluding.

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