The time period of Simple harmonic motion SHM of the mass spring system will be option 1) 2 π/3 √(m/k)
Assume that the elastic wall does not exist. The situation will be identical to the case when the wall is there. The difference is that with the wall, only a part of the SHM is executed between x = - 2e and x = -e positions. A full SHM would be between positions x = -2e, to x = -e, x = 0, x = e, to x = +2e.
Let the equation of motion be x = 2e Sin(3π/2 + 2π t/T). So at t = 0, x = - 2e, as per given condition. Let at t = t1, x = - e.
Here, T is the time period of SHM when there is no wall. Thus T = 2 π √(m/k)
so -e = 2 e Sin (3π/2 + 2π t1 /T) - 1/2 = Sin (11π/6) = Sin (3π/2 + 2 π t1 / T) 2π t1 / T = π/3 t1 = π/3 √(m/k)
The time period in the given situation is equal to = 2 * t1 = 2 π/3 * √(m/k)
This because from position x = -e the mass will rebound at the same speed due to elastic wall.
Answers & Comments
Verified answer
The time period of Simple harmonic motion SHM of the mass spring system will be option 1) 2 π/3 √(m/k)Assume that the elastic wall does not exist. The situation will be identical to the case when the wall is there. The difference is that with the wall, only a part of the SHM is executed between x = - 2e and x = -e positions. A full SHM would be between positions x = -2e, to x = -e, x = 0, x = e, to x = +2e.
Let the equation of motion be x = 2e Sin(3π/2 + 2π t/T).
So at t = 0, x = - 2e, as per given condition.
Let at t = t1, x = - e.
Here, T is the time period of SHM when there is no wall. Thus
T = 2 π √(m/k)
so -e = 2 e Sin (3π/2 + 2π t1 /T)
- 1/2 = Sin (11π/6) = Sin (3π/2 + 2 π t1 / T)
2π t1 / T = π/3
t1 = π/3 √(m/k)
The time period in the given situation is equal to = 2 * t1
= 2 π/3 * √(m/k)
This because from position x = -e the mass will rebound at the same speed due to elastic wall.