Question 2 of 14intermediate🔧 ApplyNumerical3 marks
In Exercise 13.9, let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a)
at the mean position,
Answer
x(t) = 2 sin(20t) cm
(b)
at the maximum stretched position, and
Answer
x(t) = 2 cos(20t) cm
(c)
at the maximum compressed position.
Answer
x(t) = -2 cos(20t) cm or x(t) = 2 sin(20t + 3π/2) cm
Explanation
From Exercise 13.9, we have spring constant k = 1200 N/m, mass m = 3 kg, and amplitude A = 2.0 cm. The angular frequency ω = = = = 20 rad/s. The general SHM equation is x(t) = A sin(ωt + φ), where φ is determined by initial conditions.