JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Elasticity & SHM

On a frictionless horizontal plane, a bob of mass

m = 0.1 k g

is attached to a spring with natural length

l_{0} = 0.1 m

. The spring constant is

k_{1} = 0.009 N m^{- 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 = (nπ) s

, then the integer closest to

n

is ____.

Answer: 6

Solution

The oscillation consists of two parts with different spring constants.

Time period for extension (

l > l_{0}

T_{1} = π \frac{m}{k _{1}} = π \frac{0.1}{0.009} = \frac{π}{0.3}

Time period for compression (

l < l_{0}

T_{2} = π \frac{m}{k _{2}} = π \frac{0.1}{0.016} = \frac{π}{0.4}

Total time period:

T = T_{1} + T_{2} = \frac{π}{0.3} + \frac{π}{0.4} = π (\frac{4 + 3}{1.2}) = \frac{7 π}{1.2} \approx 5.83 π \approx 6 π

Therefore,

n = 6

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer