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Grade 11Mechanics

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.

Radhika Batra

12 Years agoGrade 11

2 Answers

Kevin Nash

12 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

mω² = YA / L

⇒ ω = √ YA / mL ⇒ 2π / T = √YA / mL

Frequency f = 1 / T = 1 / 2π √ YA / mL

Thanks
Kevin Nash
askIITians Faculty

Rishi Sharma

5 Years 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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