A sonometer's fundamental note is 50Hz, what is the new frequency when the tension is four times the original?

Analyze the question A sonometer is a stretched wire (string). We're given the fundamental frequency (50 Hz) and told the tension becomes 4 times the original. We need the new frequency, assuming the length and mass per unit length stay the same.

Concept For a vibrating string, the fundamental frequency is:

f = (1/2L) × √(T/μ)

where L is the length, T is the tension and μ is the mass per unit length.

When L and μ are constant, f ∝ √T, so:

f₂/f₁ = √(T₂/T₁)

Solution

• f₁ = 50 Hz
• T₂ = 4T₁

f₂ = f₁ × √(T₂/T₁) f₂ = 50 × √4 f₂ = 50 × 2 f₂ = 100 Hz

Answer: B. 100 Hz

Tip Frequency changes with the square root of tension. So if the tension is multiplied by 4, the frequency is multiplied by 2. If the tension is multiplied by 9, the frequency is multiplied by 3. Don't multiply the frequency by 4, because that is the common trap (it would give 200 Hz, option C).