On a frictionless horizontal plane, a bob of mass is attached to a spring with natural length The spring constant is when the length of the spring and is when Initially the bob is released from Assume that Hooke's law remains valid throughout the m

Question

On a frictionless horizontal plane, a bob of mass $m = 0.1 kg$ is attached to a spring with natural length $l_{0} = 0.1 m .$ The spring constant is $k_{1} = 0.009 {Nm}^{- 1}$ when the length of the spring $l > l_{0}$ and is $k_{2} = 0.016 {N m}^{- 1}$ when $l < l_{0} .$ Initially the bob is released from $l = 0.15 m .$ Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T = (n π) s,$ then the integer closest to $n$ is _______. [2022]

Smart Achievers · Accepted Answer

(6)

$We have$

$ℓ > ℓ_{0} \Rightarrow k = k_{1}; ℓ < ℓ_{0} \Rightarrow k = k_{2}$

Time period of oscillation,

$T = π \sqrt{\frac{m}{k_{1}}} + π \sqrt{\frac{m}{k_{2}}}$ $= π \sqrt{\frac{0.1}{0.009}} + π \sqrt{\frac{0.1}{0.016}}$

$= \frac{π}{0.3} + \frac{π}{0.4}$ $= π \times \frac{0.70}{0.12}$ $= 5.83 π \approx 6 π$

So, $n = 6$

Solution

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Solution

Ans. (6) We have ℓ>ℓ0⇒k=k1 ; ℓ<ℓ0⇒k=k2 Time period of oscillation, T=πmk1+πmk2=π0.10.009+π0.10.016 =π0.3+π0.4=π×0.700.12=5.83π≈6π So, n=6

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