consider the situation shown in the figure show that the blocks are displaced slightly in opposite directions and released ,they will execute simple harmonic motion.calculate time period.
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Solution : The centre of mass of the system should not change during the motion . so if the block 'm' on the left a distance 'x' so total compression of the spring is 2x
By energy method
1/2k(2x)² + 1/2mv² +1/2mv² = C
mv² + 2kx² = C
taking darivative of both sides with respect to T
m×2v dv/dt + 2k×2x dx/dt = 0
•°•ma + 2kx = 0 (because v = dx/dt and a = dv/dt)
a/x = -2k/m => w² = w= √2k/m
time period T = √2k/m Answer ✔
By energy method
1/2k(2x)² + 1/2mv² +1/2mv² = C
mv² + 2kx² = C
taking darivative of both sides with respect to T
m×2v dv/dt + 2k×2x dx/dt = 0
•°•ma + 2kx = 0 (because v = dx/dt and a = dv/dt)
a/x = -2k/m => w² = w= √2k/m
time period T = √2k/m Answer ✔
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Explanation:
The centre of mass of the system should not change during the motion. So, if the block ‘m’ on the left moves towards right a distance ‘x’, the block on the right moves towards left a distance ‘x’. So, total compression of the spring is 2x.
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