Question

3. At t = 0, an 885-g mass at rest on the end of a horizontal spring (k = 184 N/m) is struck by a hammer which gives it an initial speed of 2.26 m/s. Determine (a) the period and frequency of the motion, (b) the amplitude, (c) the maximum acceleration, (d) the total energy, and (e) the kinetic energy when x = 0.4A where A is the amplitude.

Answer 1

(a) Period = 0.436 s, frequency = 2.29 Hz

(b) 0.157 m

(c) 32.6 m/s²

(d) 2.26 J

(e) 1.91 J

Step-by-step explanation:

This is s simple harmonic motion in the form of a loaded spring. The initial velocity is the maximum velocity because the maximum velocity occurs at the equilibrium point.

Given:

Mass, m = 885 g = 0.885 kg

Spring constant, k = 184 N/m

Maximum velocity,
`v_m` = 2.26 m/s

(a) Period, T, is given by:

`T=2\pi\sqrt{(m)/(k)}`

`T=2\pi\sqrt{\frac{0.885 \text{ kg}}{184\text{ N/m}}} = 0.436 \text{ s}`

Frequency, f, is given by

`f= (1)/(T) = \frac{1}{0.436\text{ s}} = 2.29\text{ Hz}`

(b) The maximum velocity is related to the amplitude by

`v_m=\omega A`

where A is the amplitude ω is the angular velocity and is given by

`\omega=2\pi f = \sqrt{(k)/(m)}`

`A = (v_m)/(2\pi f) = \frac{2.26\text{ m/s}}{2\pi*2.29\text{ Hz}} = 0.157\text{ m}`

(c) Maximum acceleration is given by

`a = \omega^2A = (k)/(m)A`

`a = \frac{184\text{ N/m}}{0.885\text{ kg}}*0.157\text{ m} = 32.6\text{ m/s}^2`

(d) Because the total energy is equal at any time, we determine the energy at equilibrium.

At equilibrium, displacement is 0 m and velocity is maximum. Hence, the potential energy is 0 J and the kinetic is maximum which is determined thus:

`KE_\text{max} = (1)/(2)mv_m^2 = (1)/(2)(0.885\text{ kg})(2.26 \text{ m/s})^2 = 2.26\text{ J}`

The maximum energy is 2.26 J.

(e) The velocity at any point is

`v = \omega√(A^2 - x^2)`

At point x = 4A,

`v = \omega√(A^2 - (0.4A)^2) = \omega√(0.84A^2) = \omega A√(0.84)`

Putting the values of A and ω,

`\omega = \sqrt{(k)/(m)} = 14.42\text{ rad/s}`

`v = (14.42\text{ rad/s})(\text{0.157 m})√(0.84) = 2.075\text{ m/s}`

The kinetic energy is

`KE = (1)/(2)mv^2 = (1)/(2)(0.885\text{ kg})(2.075\text{ m/s})^2 = 1.91\text{ J}`