Question

An object with mass 3.7 kg is executing simple harmonic motion, attached to a spring with spring constant 320 N/m . When the object is 0.017 m from its equilibrium position, it is moving with a speed of 0.60 m/s . Part A Calculate the amplitude of the motion. Express your answer to two significant figures and include the appropriate units. A = nothing nothing

Answer 1

Answer:

The amplitude of the motion of the spring is 1 m.

Explanation:

From the conservation of energy;

$(PE_{spring})_{max}= PE+KE$

which is

$\frac{1}{2} kx^2_{max}=\frac{1}{2} kx^2+\frac{1}{2} mv^2$

Make $x_{max}$ the subject of formula

$x^2_{max}= \frac{kx^2\times mv^2}{k}$

$x_{max}= \sqrt{\frac{kx^2\times mv^2}{k} }$

Substitute 320 N/m for k, 0.017 m for x, 3.7 kg for m, and 0.60 m/s for v.

$x_{max}= \sqrt{\frac{(320)(0.017)^2+ (3.7)(0.60)^2}{320} } \\\\=1m$

Thus, The amplitude of the motion of the spring is 1 m.

Answer 2

Answer: 0.0667m

Explanation:

given

Mass of the object, m = 3.7kg

Force constant of the spring, k = 320 N/m

Speed of the object, v = 0.6 m/s

Distance from equilibrium position, x = 0.017 m

Using the law of conservation of energy. The total energy conserved is

= 1/2kx² + 1/2mv²

= 1/2 * 320 * 0.017² + 1/2 * 3.7 * 0.6²

= 0.046 + 0.666

= 0.712 J

When v = 0, the maximum deflection is Xmas. This Xmas, is also the amplitude. Such that,

Energy = 1/2kx²

0.712 = 1/2 * 320 * x²

1.424 = 320 x²

x² = 1.424 / 320

x² = 0.00445

x = 0.0667

x = 6.67*10^-2 m

Thus, the amplitude is 0.0667 m