the time taken by a particle performing shm to pass from pt a to b where its velocities are same is 2s.after another 2 secs it returns to b.the time period of oscillation is?

Question

Subrata Dutta · Answer

According to the question, a and b points are such that they are located at same distances from the equilibrium position. Here also it is said that velocity is same ,i.e; not only the magnitude but also the directions are same. So, total time period of oscillation = 2×( time taken to go from a to b + the next time taken to return at b) = 2×(2+2)= 8 sec(ans)

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem. According to the question, a and b points are such that they are located at same distances from the equilibrium position. Here also it is said that velocity is same ,i.e; not only the magnitude but also the directions are same. So, total time period of oscillation = time taken to go from a to b + the next time taken to return at b + time taken to go from b to a + time taken to return to a = 2 x (time taken to go from a to b + the next time taken to return at b) = 2×(2+2) = 8 sec Hope it helps. Thanks and regards, Kushagra

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the time taken by a particle performing shm to pass from pt a to b where its velocities are same is 2s.after another 2 secs it returns to b.the time period of oscillation is?

the time taken by a particle performing shm to pass from pt a to b where its velocities are same is 2s.after another 2 secs it returns to b.the time period of oscillation is?

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the time taken by a particle performing shm to pass from pt a to b where its velocities are same is 2s.after another 2 secs it returns to b.the time period of oscillation is?

the time taken by a particle performing shm to pass from pt a to b where its velocities are same is 2s.after another 2 secs it returns to b.the time period of oscillation is?

2 Answers

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