simple harmonic motion questions (1 Viewer)

[paper moon]

Does anyone know how to solve these two questions:
1. Assume that over days of constant weather the cycle of temperatures each day is S.H. between 13 degrees at 4 am and 23 degrees at 4 pm. At what times of the day would the temperature be iii) 15 degrees ANS:7:33am
2. A particle states at t=0 from a position 8 m from the origin, moving away from the origin with a velocity 4 m/s. If the acceleration is a constant 2 m/s towards the origin find an expression for velocity as a function of displacement. ANS: v=+- 2 sqr(12-x)

 

[Drongoski]

Q1

amplitude = 5 degrees (a = 5)

period T = 24 hrs = 2 pi/n so that n = pi/12

x = -5cos nt (we have a SHM, centre 0, oscillating between -5 and +5 [i.e. between 13 deg & 23 deg])

when temperature = 15 deg, x = -3

Solve for: -3 = -5cos((pi x t)/12)

You get t = 3.54315 ... hrs = 3 hr 32.589 .. mins = 3 hr 33 min (approx)

Therefore 18 deg occurs at 4 am + 3 hr 33 min = 7:33 am

 

[Aesytic]

Q2:

acceleration = d(1/2*v^2)/dx
= -2
integrating with respect to x,
1/2 * v^2 = -2x + C
when t=0, x=8 and v=4
.'. 1/2 * (4^2) = -2*8 + C
16/2 = -16 + C
8 = -16 + C
C = 24
.'. 1/2 * v^2 = -2x + 24
v^2 = -4x + 48
v = +- sqrt(-4x + 48)
= +- 2sqrt(12-x)