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

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

Step-by-step explanation:

From the conservation of energy;

`(PE_(spring))_(max)= PE+KE`

which is

`(1)/(2) kx^2_(max)=(1)/(2) kx^2+(1)/(2) mv^2`

Make
`x_(max)` the subject of formula

`x^2_(max)= (kx^2* mv^2)/(k)`

`x_(max)= \sqrt{(kx^2* 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{((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

Step-by-step 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