The block of mass m shown in Fig. (a) is at rest in the equilibrium position at x = 0 when it receives an impulse that results in the initial velocity ẋ (0) = v0. (1) Derive the differential equation of motion for the block. Assuming that the system is critically damped, determine (2) the damping

Question

The block of mass m shown in Fig. (a) is at rest in the equilibrium position at x = 0 when it receives an impulse that results in the initial velocity [latex]\dot x (0) = v_0[/latex]. (1) Derive the differential equation of motion for the block. Assuming that the system is critically damped, determine (2) the damping coefficient c; and (3) the maximum displacement of the block.

Accepted Answer

Part 1

The free-body (FBD) and mass-acceleration (MAD) diagrams of the block in an arbitrary position are shown in Fig. (b). The forces acting on the mass are its weight mg, the resultant normal reaction N, the two spring forces, and the force exerted by the damper. The differential equation of motion in the x-direction is

[latex]∑F_x = ma_x \underrightarrow{+} −k_1x − k_2x − c\dot x = m\ddot x[/latex]

or

[latex]m\ddot x + c\dot x + (k_1 + k_2)x = 0[/latex] (a)

Part 2

By comparing Eq. (a) with Eq. (20.11), we deduce that the undamped circular frequency is

[latex]m\ddot x + c\dot x + kx = 0[/latex] (20.11)

[latex]p = \sqrt{\frac{k_1 + k_2}{m}}[/latex] (b)

Therefore, the critical damping coefficent is—see Eq. (20.14)—

[latex]c_{cr} = 2mp [/latex] (20.14)

[latex]c =c_{cr} = 2mp = 2m\sqrt{\frac{k_1 + k_2}{m}} = 2\sqrt{m(k_1 + k_2)}[/latex]

Part 3

The displacement of the block is given by Eq. (20.18)

[latex]x = (A_1 + A_2t)e^{−pt}[/latex] (20.18)

[latex]x = (A_1 + A_2t)e^{−pt}[/latex] (c)

Hence, the velocity is

[latex]\dot x = A_2e^{−pt} − p(A_1 + A_2t)e^{−pt}[/latex] (d)

Substituting the initial conditions x = 0 and [latex]\dot x = v_0[/latex] at t = 0 into Eqs. (c) and (d) yields [latex]A_1 = 0[/latex] and [latex]A_2 = v_0[/latex]. Consequently,

[latex]x = v_0te^{−pt} \dot x = v_0(1 − pt)e^{−pt}[/latex]

The maximum displacement occurs when [latex]\dot x[/latex] = 0, which happens at time t = 1/p. Therefore, the maximum displacement is

[latex]x_{max} = \frac{v_0}{p} e^{−1} = 0.368v_0\sqrt{\frac{m}{k_1 + k_2}}[/latex]

Question 20.3: The block of mass m shown in Fig. (a) is at rest in the equi...

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