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The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in...

The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers to three significant figures. Determine:

amplitude= 2cm

period= 0.628s

spring constant =5 N/m

phase constant= -0.785 rad

find initial coordinate of the mass and the initial velocity.

Solutions

Expert Solution

Mass of the object = m = 50 g = 0.05 kg

Amplitude of the oscillation = A

Angular frequency of the oscillation =

Phase constant of the oscillation =

Spring constant = k

X(t) = 2Cos(10t - /4)

Comparing to the general equation,

X(t) = ACos(t + )

A = 2 cm

= 10 rad/s

= -/4 = -0.785 rad

Time period of the oscillation = T

T = 2/

T = 2/10

T = 0.628 sec

² = k/m

(10)² = k/0.05

k = 5 N/m

X(t) = 2Cos(10t - /4)

V = dX/dt

Differentiating with respect to time,

V(t) = -(2x10)Sin(10t - /4)

V(t) = -20Sin(10t - /4)

For initial position and velocity we take t = 0 sec.

X(t) = 2Cos(10t - /4)

X(0) = 2Cos(10(0) - /4)

X(0) = 1.414 cm

V(t) = -20Sin(10t - /4)

V(0) = -20Sin(10(0) - /4)

V(0) = 14.14 cm/s

Amplitude = 2 cm

Period = 0.628 sec

Spring constant = 5 N/m

Phase constant = -0.785 rad

Initial coordinate of the mass = 1.414 cm

Initial velocity of the mass = 14.14 cm/s

genius_generous answered 1 year ago

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