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4 years ago

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A mass of 0.560 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic mo

tion of the mass is described by x(t) = (0.700 m)cos[(12.0 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass m (b) force constant for the spring N/m (c) position of the mass after it has been oscillating for one half a period m (d) position of the mass two-thirds of a period after it has been released m (e) time it takes the mass to get to the position x = −0.300 m after it has been released

2 answers:

Slav-nsk [51]4 years ago

8 0

Answer:

(a) A = 0.700m

(b) k = 80.6N/m

(c) x = -0.699m

(d) x = -0.350m

(e) t = 0.168s

Explanation:

Given the equation of motion for the spring

X = 0.700cos(12.0t), m = 0.56kg

(a) A = amplitude = 0.700m

(b) The angular velocity ω = 12rad/s

ω = √(k/m)

ω² = k/m

k = m×ω² = 0.56×12² = 80.6N/m

Spring constant k = 80.6N/m

(c) T = 2π/ω = 2π/12

T = 0.524s

At t = T/2 = 0.524/2 = 0.262s

So x = 0.700cos(12×0.262) = –0.699m

(d) At t = 2/3×T = 2×0.524/3 = 0. 349s

x = 0.700cos(12×0.349) = –0.350m

(e) to find t at x = -0.300m

–0.300 = 0.700cos(12t)

–0.300/0.700 = cos(12t)

cos(12t) = –0.429

12t = cos-¹(-0.429)

12t = 2.01

t = 2.01/12

t = 0.168s

Sphinxa [80]4 years ago

6 0

Given the equation of motion for the spring

X = 0.700cos(12.0t)

Mass of the body attached is

m = 0.56kg

(a) comparing the given equation to the general spring equation

X = A Cos(wt)

Where A is amplitude and w is frequency

Then, A = 0.7m

A = amplitude = 0.700m

(b) from the wave equation, we can deduced that,

The angular velocity is ω = 12rad/s

ω = √(k/m)

ω² = k/m

k = m×ω² = 0.56×12² = 80.6N/m

Spring constant k = 80.6N/m

(c) T = 2π/ω = 2π/12

T=π/6

At half period t = T/2

At t = T/2 = π/12

Since, X = 0.700cos(12.0t)

So, x = 0.700cos(12×π/12)

X = 0.700cos(π)

x = -0.7m

(d) at two-third of the period

t = ⅔ × 2π/12

t = π/9

X = 0.700cos(12.0t)

x = 0.700Cos(12×π/9)

x = 0.700Cos(4π/3)

x = 0.700 × -0.5

x = -0.35m

(e) to find t at x = -0.300m

–0.300 = 0.700cos(12t)

–0.300/0.700 = cos(12t)

cos(12t) = –0.429

12t = cos-¹(-0.429)

12t = 2.014

t = 2.01/12

t = 0.168s

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