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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
8 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
one year 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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