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A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to oscillate. What is the spring’s elastic constant?a) 3.5 N/mb) 5.2 N/mc) 1.9 N/md) 27 N/m

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hamzaahmeds

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

[tex]Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}[/tex]

[tex]\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}[/tex]

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

[tex]k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\[/tex]

k = 3.5 N/m

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