A sonometer wire of length 1.5 m is made of steel. the tension in it produce an elastic strain of 1%.What is the fundamental frequency of steel if density and elasticity of steel are 7.7x10^3 kg/m^3 and 2.2x10^11 N/m^2 respectively
Answer is 178.2Hz.
Plz help to solved out it in simple language
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Transverse waves on a Sonometer wire/bridge.
Young's modulus Y = Stress / Strain = 2.2 * 10^11 N/m²
Stress = Tension / Cross section area = T / A
Strain = 1 % = 0.01
2.2 * 10^11 = (T / A ) / 0.01 = T / (0.01 A)
T = 2.2 * 10^9 * A Newtons
L = 1.5 meter
Volume density d = 7.7 * 10³ kg/m³
Linear density μ = volume density * Area = d * A = 7.7 * 10³ * A kg /meter
velocity of transverse wave on the Sonometer wire = √(T/μ)
v = √(2.2*10^9 / 7.7*10³) m/sec
= 534.52 m/sec
Fundamental note has a wavelength = λ = 2 * L = 3 meters. As the given wire vibrates with one loop. Nodes at both ends and anti-node at the center. So half of the wavelength is equal to the length of the wire.
So frequency = v / λ = 178.17 Hz
Young's modulus Y = Stress / Strain = 2.2 * 10^11 N/m²
Stress = Tension / Cross section area = T / A
Strain = 1 % = 0.01
2.2 * 10^11 = (T / A ) / 0.01 = T / (0.01 A)
T = 2.2 * 10^9 * A Newtons
L = 1.5 meter
Volume density d = 7.7 * 10³ kg/m³
Linear density μ = volume density * Area = d * A = 7.7 * 10³ * A kg /meter
velocity of transverse wave on the Sonometer wire = √(T/μ)
v = √(2.2*10^9 / 7.7*10³) m/sec
= 534.52 m/sec
Fundamental note has a wavelength = λ = 2 * L = 3 meters. As the given wire vibrates with one loop. Nodes at both ends and anti-node at the center. So half of the wavelength is equal to the length of the wire.
So frequency = v / λ = 178.17 Hz
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