Physics, asked by rushyendra09, 7 months ago

An aluminum wire and a steel wire of same cross sectional area of 10-2 cm2 were connected together. A 10kg stone was loaded on the compound wire as shown
Waves are set up in compound wire. The minimum frequency of excitation for which the standing waves are observed with point A as node is L 60cm and
L2 = 86.6cm (in cycles/s)
7
(Linear density of aluminum wire is 2.6 unit, linear density of steel wire=7.8 units)
stone
MF10 kg

Answers

Answered by knjroopa

Explanation:

Given An aluminium wire and a steel wire of same cross sectional area of 10-2 cm^2 were connected together. A 10 kg stone was loaded on the compound wire as shown Waves are set up in compound wire

Now we have wires 1 and 2.So we need to form a node in the middle.
So the two distances are L1 and L2. So the harmonics are n1 and n2.
Also Tension T = 10 kg = 100 N, ρ = 2.6 x 10^3 kg / m^3, Area = 10^-2 x 10^-4 = 10^-6 m^2
So we have n1 x λ1/2 = L1 and n2 x λ2/2 = L2 -----------1
So velocity of wave v = f x λ
Or λ = v/f
So n1 / 2 x v1 / f = L1 and n2 x v2 / 2f = L2
So we have
n1 / n2 = L1 / L2 x v2 / v1
Now velocity v = √T/μ (T x L / Mass
T x L / ρ x A x L)
So v = √T / ρ x A
So we need to find v2/v1
So v2 / v1 = √ρ1 / ρ2
So we get
n1 / n2 = L1 / L2 x √ρ1 / ρ2
n1 / n2 = 60 / 86.6 x √2.6 / 7.8
= 2/5
Now n1 = 2 and n2 = 5
So in wire 1 it will be 2nd harmonic and in wire 2 it will be 5th harmonic.
So there are 2 loops and 5 loops in wire 1 and 2.
So substituting n1 and n2 values in equation 1 we get
2 x λ1/2 = L1
λ1 = L1
v/f = L1
So f = v/L1
We can write this as
f = 1/L1 √T / ρ x A
= 1/0.6 √100 / 2.6 x 10^3 x 10^-6
= 200 / 0.6
= 333 hz