The fundamental frequency of vibration of a string stretched betw… — Solution | JEE Physics

NVAEasyJEE 2023Superposition Principle & Standing Waves

The fundamental frequency of vibration of a string stretched between two rigid supports is $50\,Hz$ . The mass of the string is $18\,g$ and its linear mass density is $20\,g\,m^{-1}$ . The speed of the transverse waves produced in the string is _____ $m\,s^{-1}$ .

Answer

Correct answer:90

Step-by-step solution

Standard Method

Given: fundamental frequency $f = 50\,Hz$ , mass of string $m = 18\,g$ , linear mass density $\mu = 20\,g\,m^{-1}$ .

Find: speed of transverse waves $v$ .

From linear mass density,

\mu = \frac{m}{L}

So,

L = \frac{m}{\mu} = \frac{18}{20} = 0.9\,m

For a string fixed at both ends, the fundamental frequency is

f = \frac{v}{2L}

Substitute the given values:

50 = \frac{v}{2 \times 0.9}

Hence,

v = 2 \times 0.9 \times 50 = 90\,m\,s^{-1}

Therefore, the speed of the transverse waves in the string is $90\,m\,s^{-1}$ .

Direct Relation

Given: $m = 18\,g$ , $\mu = 20\,g\,m^{-1}$ , $f = 50\,Hz$ .

Find: $v$ .

First compute the length quickly using

L = \frac{m}{\mu} = \frac{18}{20} = 0.9\,m

Then use the fundamental mode relation

v = 2Lf = 2 \times 0.9 \times 50 = 90\,m\,s^{-1}

Therefore, the correct numerical value is 90.

Common mistakes

Using $f = \frac{v}{L}$ instead of $f = \frac{v}{2L}$ for a string fixed at both ends is incorrect because the fundamental mode has wavelength $\lambda = 2L$ . Always use $f = \frac{v}{2L}$ for the first harmonic.
Confusing total mass with linear mass density and substituting $18$ directly into the wave formula is wrong because the string length must first be found from $\mu = \frac{m}{L}$ . Compute $L$ before finding $v$ .
Making a unit-handling error while finding length can lead to a wrong result. Since both $m$ and $\mu$ are given in grams and grams per metre, their ratio directly gives length in metres, so $L = 0.9\,m$ .

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