A 2.2 kg object is attached to a horizontal spring of force constant k = 4.5 kN/

Question

A 2.2 kg object is attached to a horizontal spring of force constant k = 4.5 kN/m. The spring is stretched 10 cm from equilibrium and released. Find the frequency of the motion. Find the period. Find the amplitude. Find the maximum speed. Find the maximum acceleration. When does the object first reach its equilibrium position? What is its acceleration at this time? The frequency of the object's motion is determined by the mass and force constant. The maximum velocity and acceleration of an object executing simple harmonic motion can be determined by taking successive time derivatives of the position function.

Explanation / Answer

(a) Mass of the object, m = 2.2 kg

Spring constant, k = 4.5*1000 = 4500 N/m

Stretched distance, A = 0.10 m

Frequency, f = (1/2*pi)sqrt(k/m) = (1/2*3.14)sqrt(4500/2.2) = 7.20 Hz

(b) Time period, T = 1/f = 1/7.20 = 0.139 s

(c) Amplitude = A = 0.10 m

(d) Maximum speed = Aw = A*(2*pi*f) = 0.10(2*3.141*7.20) = 4.52 m/s.

(e) Maximum acceleration = A*w^2 = A*(2*pi*f)^2 = 0.10*(2*3.14*7.2)^2 = 204.4 m/s^2

(f) Time taken by the object to reach its equilibrium position,

t = T/4 = 0.139/4 = 0.035 s = 35 ms

Acceleration at this position = 0 m/s^2

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