A block is attached a horizontal spring (k = 12.4 N/m) exhibits simple harmonic

Question

A block is attached a horizontal spring (k = 12.4 N/m) exhibits simple harmonic motion on a flat, frictionless table. The mass is observed to have a maximum displacement of 8.50 cm from its equilibrium position and completes 8 cycles in 10 seconds. At time t = 0, the mass is 1.00 cm to the left of equilibrium and is travelling to the left. Assume the positive direction is to the right. Find the angular frequency of the block. Find the mass of the block. Write down the equation of motion in the form x(t) = A cos (omega t + phi_0), using appropriate units for all constants. Assume x = 0 represents when the block is at its equilibrium position. Find the total energy stored in the block at t = 3.0 s.

Explanation / Answer

Here .

k = 12.4 N/m

amplitude , A = 8.5 cm = 0.085 m

Time period , T = 10/8 = 1.25 s

a)

w = 2pi/T

w = 2pi/1.25

w = 5.03 rad/s

the angular frequency of the block is 5.03 rad/s

b)

let the mass of block is m

as w = sqrt(k/m)

5.03 = sqrt(12.4/m)

m = 0.491 Kg

the mass of block is 0.491 Kg

c)

Here ,

as x = A * cos(w * t + phi)

at t = 0

- 1 = 8.5 * cos(w * 0 + phi)

solving

phi= 96.8 degree = 1.69 rad

the equation is

x = 8.5 * cos(5.03 * t + 1.69) cm

d)

as total energy is conserved

Total energy stored in the block = 0.5 * k * A^2

Total energy stored in the block = 0.5 * 12.4 * 0.085^2

Total energy stored in the block = 0.045 J

the Total energy stored in the block is 0.045 J

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