A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.70 N is applied. A 0.550-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to 5.00 cm and released from rest at t 0. Assume that the direction of the initial displacement is positive.) (a) What is the force constant of the spring? N/m (b) What are the angular frequency (c), the frequency, and the period of the motion? rad/s Hz (c) What is the total energy of the system? (d) What is the amplitude of the motion? Cm (e) What are the maximum velocity and the maximum acceleration of the particle? m/s maX m/s amax. f) Determine the displacement x of the particle from the equilibrium position at t 0.500 s Cm (g) Determine the velocity and acceleration of the particle when t 0.500 s. (Indicate the direction with the sign of your answer.) m/s m/s

Explanation / Answer

a)

We know,

F = k*x

where k = Spring constant

x = displacement = 3 cm = 0.03 m

So, 8.7 = k*0.03

So, k = 290 N/m <-------answer

b)

We know,

W = sqrt(k/m)

where W = angular frequency

So, W = ssqrt(290/0.55) = 22.96 rad/s <-----answer

Now, f = W/(2*pi) = 22.96/(2*pi) = 3.65 Hz <--------answer

So, T = 1/f = 0.274 s <------answer

c)

Total energy,Umax = 0.5*k*X^2

where X = 5 cm = 0.05 m

So, Umax = 0.5*290*0.05^2 = 0.363 J <----------answer

d)

amplitude = displacement = 5 cm <-------answer

e)

For max velocity,Vmax,

we can use energy conservation,

So KE = PE

So, 0,5*m*Vmax^2 = 0.5*k*X^2

So, 0.55*Vmax^2 = 290*0.05^2

So, Vmax = 1.15 m/s <-------answer

a_max = k*X/m = 290*0.05/0.55 = 26.4 m/s2 <------answer

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