One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass (m) hangs freely from the free end of the spring. The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released it will oscillate with a time period T equal to

# One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass (m) hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released it will oscillate with a time period T equal to

1. A

$2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

2. B

$2\mathrm{\pi }\sqrt{\frac{\mathrm{m}\left(\mathrm{YA}+\mathrm{KL}\right)}{\mathrm{YAK}}}$

3. C

$2\mathrm{\pi }\sqrt{\frac{\mathrm{mYA}}{\mathrm{kL}}}$

4. D

$2\mathrm{\pi }\sqrt{\frac{\mathrm{mL}}{\mathrm{YA}}}$

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

As the spring is connected in series with the rod, ${\mathrm{K}}_{\mathrm{eff}}=\frac{{\mathrm{K}}_{1}{\mathrm{K}}_{2}}{{\mathrm{K}}_{1}+{\mathrm{K}}_{2}}$

${\mathrm{K}}_{\mathrm{eff}}=\frac{\frac{\mathrm{YA}}{\mathrm{L}}\mathrm{K}.}{\frac{\mathrm{YA}}{\mathrm{L}}+\mathrm{K}}=\frac{\frac{\mathrm{YAK}}{\mathrm{L}}}{\frac{\mathrm{YA}+\mathrm{LK}}{\mathrm{L}}}=\frac{\mathrm{YAK}}{\mathrm{YA}+\mathrm{LK}}$

## Related content

 Difference Between Mass and Weight Differences & Comparisons Articles in Physics Important Topic of Physics: Reynolds Number Distance Speed Time Formula Refractive Index Formula Mass Formula Electric Current Formula Electric Power Formula Resistivity Formula Weight Formula  +91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)