One end of a massless spring of relaxed length 50 cm and spring constant k is fixed on top of a frictionless inclined plane of inclination $θ = 30 °$ as shown in Fig. When a mass m = 1.5 kg is attached at the other end, the spring extends by 2.5 cm. The mass is displaced slightly and released. The time period (in seconds) of the resulting oscillation will be

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$\frac{π}{5}$

$\frac{π}{7}$

$\frac{2 π}{7}$

$\frac{2 π}{5}$

answer is A.

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Detailed Solution

The force which increases the length of the spring by x = 2.5 cm is $F = m g \sin θ$ .

Therefore, the spring constant is
$k = \frac{F}{x} = \frac{m g \sin θ}{x} Now time period T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{m}{m g \sin θ / x}} = 2 π \sqrt{\frac{x}{g \sin θ}} Putting x = 2.5 cm = 2.5 \times 10^{- 2} m, g = 9.8 {ms}^{- 2} and θ = 30 °, we get T = π / 7 second.$