a 0.03kg mass is tethered by two identical springs of force constant 2.5N/m . if the mass is now displaced by 20mm to the left to its equilibrium position and released calculate
(a) the time period and frequency of subsequent oscillations
(b) the acceleration at centre and at extremities of the oscillation
Answers
Answer :
Let the mass m be displaced by a small distance x to the left from its mean position as shown in second attachment.
Due to it the spring on the right side gets stretched by a length x while that on the left side gets compressed by the same length
The forces acting on the mass are
- F₁ = -kx towards left side
- F₂ = -x towards left side
The net force acting on the mass is,
- F = F₁ + F₂ = -2kx
Here, F ∝ x and -ve sign shows that force is towards the mean position, therefore the motion executed by the block is simple harmonic.
★ Acceleration is
- a = F/m = -2kx/m
The standard equation of SHM is
- a = -ω²x
Comparing both equations, we get
➠ ω² = 2k/m
➠ ω = √2k/m
Time period, T = 2π/ω = 2π√m/2k
➙ T = 2π√m/2k
➙ T = 2(3.14)√0.03/2(2.5)
➙ T = 6.28√0.006
➙ T = 6.28 × 0.077
➙ T = 0.48 s
We know that frequency is reciprocal of time time period.
➙ f = 1/T
➙ f = 1/0.48
➙ f = 2.08 Hz
★ Acceleration of block :
1. At centre (mean position)
➠ a = ω²x
At mean position, x = 0
➠ a = 0 m/s²
2. At extreme position :
➠ a = ω²x
At extreme position, x = 0.02m
- ω = √2k/m = 12.9
➠ a = (12.9)² × 0.02
➠ a = 3.32m/s²
