Q86 / 90 · JEE Main 2023 (13 Apr Shift 2) · Physics

In an experiment with sonometer when a mass of 180 g is attached to the string, it vibrates with fundamental frequency of 30 Hz. When a mass m is attached, the string vibrates with fundamental frequency of 50 Hz. The value of m is ______ g.

Correct answer: 500

Solution

The formula to calculate the fundamental frequency in a stretched string is given by

f=12LTμ   ...1

For the first mass, it can be written that

f1=12Lm1gμ   ...2

And, for the second mass, it can be written that

f2=12Lm2gμ   ...3

Divide equation (3) by equation (2) and simplify to obtain the required mass.

f2f1=12Lm2gμ12Lm1gμ= m2m1⇒m1m2=f1f22⇒m2=m1f2f12   ...4

Substitute the values of the known parameters into equation (4) to calculate the required mass.

m2=180 g×50 Hz30 Hz2= 500 g

Hence, the correct answer is 500.

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