Simple harmonic motion question (1 Viewer)

[ghostman_on_3rd]

hi im having trouble with this question, i need help.

A point moves with SHM; if when at distances of 3 and 4cm from the centre of its path, its velocities are 8 and 6cm/s respectively, find its period and its acceleration when at its greatest distance from the centre.

thanks in advance<!-- google_ad_section_end --></SPAN>

 

[Drongoski]

Not necessarily the best method.

 

[ghostman_on_3rd]

nice ! i understand, thanks man

 

[Joel8945]

hi im having trouble with this question, i need help.

A point moves with SHM; if when at distances of 3 and 4cm from the centre of its path, its velocities are 8 and 6cm/s respectively, find its period and its acceleration when at its greatest distance from the centre.

thanks in advance<!-- google_ad_section_end --></SPAN>

E = 0.5(m*v^2 + kx^2)

E = 0.5(0.0064m + 0.0009k)
E = 0.5(0.0036m + 0.0016k)

0.0064m + 0.0009k = 0.0036m + 0.0016k

0.0007k = 0.0028m

k = 4m

ω = 2π/T (T = period)

ω = sqrt(k/m) = sqrt(4m/m) = 2

2 = 2π/T => T = π

The period is π seconds.

When its at its greatest distance from the centre:

E = 0.5kx^2

4m*x^2 = 0.0036m + 0.0016*4m

4mx^2 = 0.01m

x^2 = 0.0025

x = 0.05m

a(t) = -ω^2*x(t) = -4*0.05 = -4*0.05 = -0.2m/s^2

so the acceleration at its farthest point from the centre is -20cm/s^2 and the period is constant throughout the SHM which was determined to be approximately 3.14s.

 

[tommykins]

rofl yeah use physics methods righto.

ignore the above post guys.