Question

Two wires of the same material and
cross-section are stretched on a sonometer
One wire is loaded with 6 kg. while the other
is loaded with 1.5 kg. If both the wires vibrate
with same fundamental frequency calculate
the vibrating length of the first wire. The
vibrating length of the second wire is 30 cm
​

Answer 1

Answer:

120

Explanation:

As we know

Fm proportional to root 1/u

u=m/l

Fm1=Fm2

L1/M1=L2/M2

30/1.5 = X/6

X = 120

Answer 2

Given : Two wires of the same material and cross-section are stretched on a sonometer One wire is loaded with 6 kg. while the other is loaded with 1.5 kg. If both the wires vibrate with same fundamental frequency.

To find : the vibrating length of the first wire if the vibrating length of the second wire is 30 cm.

solution : fundamental frequency, f = 1/2l √{T/μ}

where l is the length, μ is linear mass density and T is the tension on a wire.

here, m₁ = 6kg , m₂ = 1.5 kg

so, T₁/T₂ = 6/1.5 = 4 ......(1)

now f₁ = 1/2l₁√{T₁/μ₁}

and f₂ = 1/2l₂√{T₂/μ₂}

so, f₁/f₂ = (l₂/l₁) × √{(T₁/T₂)(μ₂/μ₁)}

same fundamental frequencies are same. f₁/f₂ = 1

⇒1 = (30cm)/(l₁) × √{(4) × (m₂/l₂)/(m₁/l₁)}

⇒1 = 30/l₁ × √{4 × 1/4 × (l₁/30)}

⇒l₁ = 30cm

Therefore the vibrating length of the first wire is 30cm.

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