Question

A 20 cm long rubber string obeys Hook's law. Initially when it is stretched to make its total length 24 cm, the lowest frequency of resonance is v₀. It is further stretched to make its total length 26 cm. The lowest frequency of resonance will now be

• A. The same as v₀
• B. Greater than v₀
• C. Lower than v₀
• D. Not possible to decide

Answer 1

Answer: (B)

Greater than v₀

Answer 2

v₀ = $\frac{1}{2L}\sqrt{\frac{T}{\mu}}$ and v₀'=- $\frac{1}{2L'}\sqrt{\frac{T}{\mu'}}$

Initially

L = 24cm,T = kx = k (4cm);

μ = $\frac{m}{(26cm)}$ (m is mass of the string) and L'=26 cm; T'=k(6 cm); μ' = $\frac{m}{26cm}$

$\frac{V_{0}^{'}}{V_{0}} = \frac{L}{L'}\sqrt{\frac{T'}{T}.\frac{\mu}{\mu'}} = \frac{24cm}{26cm}\sqrt{\frac{k(6cm)}{k(4cm)}\frac{m(24cm)}{m(26cm)}}$

thus $\frac{v_{0}^{'}}{v_{0}}$ >1 or v₀' >v₀