A wire resonates to a tunning fork of frequency 500 Hz at 30°C. Under the same load, when the wire is heated to 80°C, it provides 3 beats in 20 sec.  Find the linear coefficient of expansion of the material of the wire.

Given:

• Initial frequency f_⁰=500 Hz at 30°C
• Change in temperature ΔT=80°C−30°C=50°C
• Number of beats = 3 beats in 20 seconds, so the beat frequency Δf = 3/20 = 0.15 Hz.

We know that the change in frequency due to thermal expansion is related to the coefficient of linear expansion by the following relationship:

 Δf = f₀ × α × ΔT

Where:

• Δf is the change in frequency
• f₀ is the initial frequency
• α is the linear coefficient of expansion
• ΔT is the change in temperature

Substituting the known values:

 0.15 = 500 × α × 50
 α = 0.15 / (500 × 50)
 α = 0.15 / 25000
 α = 0.000006 (°C)⁻¹