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A point mass m is suspended at the end of a mass less wire of length l and cross section A. If Y is the Young’s modulus for the wire, obtain the frequency of oscillation for the simple harmonic motion along the vertical line.

A point mass m is suspended at the end of a mass less wire of length l and cross section A. If Y is the Young’s modulus for the wire, obtain the frequency of oscillation for the simple harmonic motion along the vertical line. 

Grade:11

2 Answers

Kevin Nash
askIITians Faculty 332 Points
6 years ago
Hello Student,
Please find the answer to your question
From fig. (b), due to equilibrium
T = mg ….(i)
But Y = T /A /ℓ / L
⇒ T = YA ℓ / L
From (i) and (ii)
mg = YA ℓ / L (iii)
From fig. (c)
Restoring force
= - [T - mg] = - [YA (ℓ + x ) / L - YA ℓ / L] [from (iii)]
= - YAx /L
On comparing this equation with F = - mω2x, we get
2 = YA / L
⇒ ω = √ YA / mL ⇒ 2π / T = √YA / mL
Frequency f = 1 / T = 1 / 2π √ YA / mL
Thanks
Kevin Nash
askIITians Faculty
Rishi Sharma
askIITians Faculty 646 Points
9 months ago
Dear Student,
Please find below the solution to your problem.

Frequency depends on spring factor and inertia factor.
In this case, stress =mg/A​
Strain =l/L​ (where l is extension)
Now, Young's modulus Y is given by
Y=stress/strain​ = (mg/A)/(l/L)​
mg = YAl/L
So, kl = YAl/L ​(∵mg=kl) (k is force constant)
∴k = YA/L​
Now, frequency is given by
n= (1/2π)*(​√k/m)​
=(1/2π)*(​√YA/Lm)

Thanks and Regards

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