Asimple harmonic oscillator consists of a block of mass 4.10 kg attached to a spring of spring constant 240 n/m. when t = 1.70 s, the position and velocity of the block are x = 0.165 m and v = 3.780 m/s.

(a) what is the amplitude of the oscillations?

what were the (b) position and (c) velocity of the block at t = 0 s?

(a) the amplitude of the oscillations is 0.52 m

(b) position of the block at t = 0 s is - 0.06 m

(c) the velocity of the block at t = 0 s is 3.96 m/s

Explanation:

given information:

m = 4.10 kg

k = 240 N/m

t = 1.70 s

x = 0.165 m

v  = 3.780 m/s

(a) the amplitude of the oscillations

x(t) = A cos (ωt+φ)

v(t) = dx(t)/dt

     =  - ω A sin (ωt+φ)

ω =

   =

   = 7.65 rad/s

v(t)/x(t) = - ω A sin (ωt+φ)/A cos (ωt+φ)

v(t)/x(t) = - ω sin (ωt+φ)/cos (ωt+φ)

v(t)/x(t) = - ω tan (ωt+φ)

(ωt+φ) = (-v/ωx)

           = (-3.780/(7.65)(0.165))

           = - 1.25

(ωt+φ) = - 1.25

φ = - 1.25 - ωt

   = - 1.25 - (7.65 x 1.70)

   = - 14.26

x(t) = A cos (ωt+φ)

A = x(t) / cos (ωt+φ)

   = 0.165/cos(-1.25)

   = 0.52 m

(b) position, at t=0

x(t) = A cos (ωt+φ)

x(0) = A cos (ω(0)+φ)

x(0) = A cos (φ)

      = 0.52 cos (-14.26)

      = - 0.06 m

(c) the velocity, at t=0

v(t) = - ω A sin (ωt+φ)

v(0) = - ω A sin (ω(0)+φ)

      = - ω A sin (φ)

      = - (7.65)(0.52) sin (-14.26)

      = 3.96 m/s

the answer would be c

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day