Physics Numerical Problems No. 8.4, Unit-8: Waves, HSSC-I (class 11) Physics (Punjab Curriculum Textbook Board, Lahore) With Verified Answers

The frequency of the note emitted by a stretched string is 300 Hz. What is the frequency of this note when: 

(a) Length of the wave is reduced by one third without changing tension.

(b) The tension is increased by one-third without changing the length of the wire. 

(Ans: 450 Hz, 346 Hz)

Data Given:

Frequency of note = f = 300 Hz

To Find:

(a) Frequency of note = f₁ = ? {When length of the wave is reduced by one third without changing tension.}

(b) Frequency of note = f₂ = ? {When the tension is increased by one-third without changing the length of the wire. }

Solution:

(a) Frequency of note = f₁ = ? {When length of the wave is reduced by one third without changing tension.}

According to given conditions let original wavelength λ is reduced by `\frac {1}{3}` then the new wavelength λ₁ will be 

λ₁ = λ - `\frac {λ}{3}` = `\frac {2λ}{3}`

As given there is no change occur in the tension F then the speed will remains constant.

v₁ = v

or

f₁ λ₁ = f λ

or

f₁ = `\frac {f λ}{λ₁}`

putting value of λ₁ = `\frac {2λ}{3}`

f₁ = `\frac {f λ}{{2λ}/3}`

or

f₁ = `\frac {3f }{2}`

putting the value of f = 300 Hz (given)

f₁ = `\frac {3 ｘ300 Hz }{2}`

f₁ = 450 Hz  -----------------Ans.

Hence the new frequency will increase to 450 Hz

(b) Frequency of note = f₂ = ? {When the tension is increased by one-third without changing the length of the wire. }

According to given conditions let the tension F is increased by `\frac {1}{3}` then the new tension F₂ will be 

F₂ = F + `\frac {F}{3}` = `\frac {4F}{3}`

Formula for fundamental frequency 

f = `\frac {1}{3L}` `\sqrt frac {F}{m}`  ----------(1)

and the new frequency after increasing tension F₂ = `\frac {4F}{3}`

f₂ = `\frac {1}{3L}` `\sqrt frac {F₂}{m}`

or

f₂ = `\frac {1}{3L}` `\sqrt frac {4F}{3m}` -------(2)

Now dividing equation (2) by (1) we will get

`\frac {f₂}{f}` = `\sqrt frac {4}{3}`

`\frac {f₂}{f}` = 1.155

or

f₂ = 1.155 ｘ f

putting f = 300 Hz (given)

f₂ = 1.1547 ｘ 300 Hz

f₂ = 346.41 Hz

or

f₂ = 346 Hz -----------------Ans. (2)

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