First slide

Applications of SHM

Question

When a mass of 0.5 kg is suspended from the free end of spring it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period? $(g = 10 m s^{- 2})$

Moderate

A

0.628 seconds
B

2.42 seconds
C

3.63 seconds
D

1.63 seonds

Solution

When 0.5 kg is suspended, extension of the spring is 0.2 m.
Since in equilibrium mg=kx, substitute relevant values
$0 .5 \times 10 = k \times 0.2 k=25 N/m$
k is force constant
Time period of spring block system, when m= 0.25 kg mass is attached
$T=2π \sqrt{\frac{m}{K}} =2π \sqrt{\frac{0.25}{25}} =2π \sqrt{\frac{25}{2500}} = \frac{2×3.14}{10} T=0.628 seconds$

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