A block of mass m connected by a spring is kept on a horizontal frictionless floor. When the spring is at its natural length l0, the block is at a distance âaâ from the wall. Now, the block is at a distance â2aâ away from the wall and released. If the collision between the block and wall is elastic, determine the time period of oscillation.
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Hey mate,
● Answer-
T = 2π √(m/k)
● Explanation-
When spring is stretched, extension in spring x = 2a-a = a.
Here, restoring force balances the SHM force,
maw^2 - kx = 0
w^2 = ka / ma
w = √(k/m)
Time period is given by-
T = 2π/w
T = 2π √(m/k)
Hope this is useful...
● Answer-
T = 2π √(m/k)
● Explanation-
When spring is stretched, extension in spring x = 2a-a = a.
Here, restoring force balances the SHM force,
maw^2 - kx = 0
w^2 = ka / ma
w = √(k/m)
Time period is given by-
T = 2π/w
T = 2π √(m/k)
Hope this is useful...
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