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My Doubts: Elastic modulus should be in $N/m^2$ but here it is in Newtons only is this correct or they just mean here $k=30 N/m$ ? What do you want to say about $\lambda = 30 N$ ?

Do you think my equations are correct to solve for correct answer

My Try:

Let $x$ be the elongation, so we have $$T\cos\theta=mg$$ $$T\sin\theta=m\omega^2 r \;\;\;,\;\;\;r=(l+x)\sin\theta$$ $$T=kx$$

Using these three equations I can solve for $\theta\;\;,\;\; T$.

Is this correct ?

  • 1
    $\begingroup$ very good that you wrote $r=l+x \sin \theta$ , if it was me , I'd have completely missed that lol. However, the equations look ok to me. $\endgroup$
    – Buraian
    Aug 5 '21 at 8:28
  • $\begingroup$ thanks , i just want to know about lambda i think it is not standard notation $\endgroup$ Aug 5 '21 at 8:29

I suspect that the author defines elastic modulus here as $$\frac{\text{tension in the string}}{\text{relative extension}}\\=\text{tension in string}\times\frac{\text{original length}}{\text{elongation}},$$ which has units Newtons.

This replaces your third equation; everything else looks fine.

  • $\begingroup$ not sure if that is the definition, have you seen such definitions in the books ? If yes then it really replaces the third equation, thanks $\endgroup$ Aug 5 '21 at 12:21
  • $\begingroup$ yes i clicked that , let me check this one too $\endgroup$ Aug 5 '21 at 12:34
  • $\begingroup$ amazing this confirms.... thank you $\endgroup$ Aug 5 '21 at 12:35

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