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One end of a long metallic wire of length $\mathrm{L}$ tied to the ceiling. The other end is tied with a massless spring of spring constant $\mathrm{K}$. Amass hangs freely from the free end of the spring. The area of cross section and the young's modulus of the wire are A and Y respectively. If the mass slightly pulled down and released, it will oscillate with a time period T equal to:
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Verified Answer
The correct answer is:
$2 \pi \sqrt{\mathrm{m}(\mathrm{YA}+\mathrm{KL}) /(\mathrm{YAK})}$
Step 1: Calculation of spring constant of wire
Young's modulus of wire, \(\mathrm{Y}=\frac{\Delta \mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}} \times \frac{\mathrm{L}}{\mathrm{A}}\)
Due to elasticity of the wire, it behaves as equivalent spring \(\left(\mathrm{K}_{\text {spring }}=\mathrm{F} / \Delta \mathrm{x}\right)\).
Hence, Spring constant for Wire, \(\mathrm{K}_{\mathrm{w}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}}=\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}\)
Step 2: Calculation of Equivalent spring constant Let a force \(\mathrm{F}\) be applied to the end of the spring. The wire will extend.
Total extension \(\Delta \mathrm{S}=\mathrm{x}_{\mathrm{wire}}+\mathrm{x}_{\text {spring }}=\frac{\mathrm{F}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{w}}}+\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{K}}\)
But \(\mathrm{F}_{\mathrm{S}}=\mathrm{F}_{\mathrm{w}} \quad\) (By Newton's, third Law, Both will apply equal and opposite forces at their joint A )
\(\therefore \Delta \mathrm{S}=\mathrm{F}_{\mathrm{w}}\left(\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}\right)\)
So, Equivalent spring constant \(\left(\mathrm{K}_{\text {eq }}\right)\) is given by:
\(\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{F}_{\mathrm{eq}}}{\mathrm{x}_{\mathrm{eq}}}=\frac{\mathrm{F}_{\mathrm{w}}}{\Delta \mathrm{S}}=\frac{1}{\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}}\)
\(\mathrm{K}_{\mathrm{eq}}=\frac{(\mathrm{K})\left(\mathrm{K}_{\mathrm{w}}\right)}{\mathrm{K}_{\mathrm{w}}+\mathrm{K}}=\frac{\mathrm{KY} \frac{\mathrm{A}}{\mathrm{L}}}{\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}+\mathrm{K}}=\frac{\mathrm{KYA}}{\mathrm{YA}+\mathrm{KL}}\)
Step 3: Calculation of Time period
Angular frequency \(\omega\) for spring block system is given by:
\(\omega=\sqrt{\frac{\mathrm{K}_{\text {eq }}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{KYA}}{\mathrm{m}(\mathrm{YA}+\mathrm{KL})}}\)
\(\therefore\) Time period, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{(\mathrm{YA}+\mathrm{KL}) \mathrm{m}}{\mathrm{KYA}}}\)
Hence, Option B is correct
Young's modulus of wire, \(\mathrm{Y}=\frac{\Delta \mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}} \times \frac{\mathrm{L}}{\mathrm{A}}\)
Due to elasticity of the wire, it behaves as equivalent spring \(\left(\mathrm{K}_{\text {spring }}=\mathrm{F} / \Delta \mathrm{x}\right)\).
Hence, Spring constant for Wire, \(\mathrm{K}_{\mathrm{w}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}}=\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}\)
Step 2: Calculation of Equivalent spring constant Let a force \(\mathrm{F}\) be applied to the end of the spring. The wire will extend.
Total extension \(\Delta \mathrm{S}=\mathrm{x}_{\mathrm{wire}}+\mathrm{x}_{\text {spring }}=\frac{\mathrm{F}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{w}}}+\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{K}}\)
But \(\mathrm{F}_{\mathrm{S}}=\mathrm{F}_{\mathrm{w}} \quad\) (By Newton's, third Law, Both will apply equal and opposite forces at their joint A )
\(\therefore \Delta \mathrm{S}=\mathrm{F}_{\mathrm{w}}\left(\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}\right)\)
So, Equivalent spring constant \(\left(\mathrm{K}_{\text {eq }}\right)\) is given by:
\(\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{F}_{\mathrm{eq}}}{\mathrm{x}_{\mathrm{eq}}}=\frac{\mathrm{F}_{\mathrm{w}}}{\Delta \mathrm{S}}=\frac{1}{\frac{1}{\mathrm{~K}_{\mathrm{w}}}+\frac{1}{\mathrm{~K}}}\)
\(\mathrm{K}_{\mathrm{eq}}=\frac{(\mathrm{K})\left(\mathrm{K}_{\mathrm{w}}\right)}{\mathrm{K}_{\mathrm{w}}+\mathrm{K}}=\frac{\mathrm{KY} \frac{\mathrm{A}}{\mathrm{L}}}{\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}+\mathrm{K}}=\frac{\mathrm{KYA}}{\mathrm{YA}+\mathrm{KL}}\)
Step 3: Calculation of Time period
Angular frequency \(\omega\) for spring block system is given by:
\(\omega=\sqrt{\frac{\mathrm{K}_{\text {eq }}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{KYA}}{\mathrm{m}(\mathrm{YA}+\mathrm{KL})}}\)
\(\therefore\) Time period, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{(\mathrm{YA}+\mathrm{KL}) \mathrm{m}}{\mathrm{KYA}}}\)
Hence, Option B is correct
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