Question

The period of an oscillating particle is 56 s, and its amplitude is 18 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals. (a) t = 0 to t = 14 s cm (b) t = 14 s to t = 28 s cm (c) t = 0 to t = 7 s cm (d) t = 7 s to t = 14 s cm

Answer 1

a) interval t=0 to t =14 s = T/4

given, T = 56 s

in oscillation,

x = A*sin(w*t)

here, t = T/4

A = amplitude = 18 cm

w = 2*pi/T

x = 18*sin((2*pi/T)*(T/4)) = 18

So, after time period 't' it will be at first peak.

then, distance travelled = d = 18 cm

b) interval t=0 to t =14 s = another T/4 period, from the top back to the equilibrium position

Similarly in that time it reaches at equilibrium position.

So, distance travelled = d = 18 cm

c)

interval t=0 to t =7 s = T/8

in oscillation,

x = A*sin(w*t)

here, t = T/8

So, x = 18*sin((2*pi/T)*(T/8))

x = position of particle = 12.73 cm

then, distance = d = 12.73 cm

d)

interval t=7 to t =14 s = another T/8

from above answers,

distance = d = 18 - 12.73

d = 5.27 cm

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