Question 6 (paper Question VI): 04-BS-3, December 2014, Q6

Question 6 (paper Question VI) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $m=25\ \text{kg}$; springs $k_1=5\ \text{kN/m}$ and $k_2=20\ \text{kN/m}$ in series; that series pair in parallel with $k_3=2\ \text{kN/m}$; amplitude $X=25\ \text{mm}=0.025$ m.

Find. Period $\tau$, frequency $f$, and $v_{\max}$, $a_{\max}$.

Figure 6 — block on a series/parallel spring combination.

Approach. Reduce the spring network to a single effective stiffness (series, then parallel), form the natural frequency $\omega_n=\sqrt{k_{\text{eff}}/m}$ for simple harmonic motion, then use the SHM amplitude relations for peak speed and acceleration.

Effective stiffness. $$k_{\text{series}} = \frac{k_1k_2}{k_1+k_2} = \frac{5(20)}{5+20} = 4\ \text{kN/m}$$ $$k_{\text{eff}} = k_{\text{series}}+k_3 = 4+2 = 6\ \text{kN/m} = 6000\ \text{N/m}$$
Natural frequency, period. $$\omega_n = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{6000}{25}} = 15.49\ \text{rad/s}$$ $$f = \frac{\omega_n}{2\pi} = 2.466\ \text{Hz}, \qquad \tau=\frac1f=0.4056\ \text{s}$$ $$\boxed{\tau = 0.406\ \text{s}, \qquad f = 2.47\ \text{Hz}}$$
Peak velocity and acceleration (SHM: $v_{\max}=\omega_n X$, $a_{\max}=\omega_n^2 X$, with $X=0.025$ m): $$v_{\max} = 15.49(0.025) = 0.3873\ \text{m/s}$$ $$a_{\max} = 15.49^2(0.025) = 240(0.025) = 6.00\ \text{m/s}^2$$ $$\boxed{v_{\max} = 0.387\ \text{m/s}, \qquad a_{\max} = 6.00\ \text{m/s}^2}$$

Simple harmonic motion of the 25 kg block
Quantity	Value
Effective stiffness, $k_{\text{eff}}$	6.00 kN/m
Natural (circular) frequency, $\omega_n$	15.49 rad/s
Frequency, $f$	2.47 Hz
Period, $\tau$	0.406 s
Maximum velocity, $v_{\max}$	0.387 m/s
Maximum acceleration, $a_{\max}$	6.00 m/s²

Topic: Free undamped vibration (simple harmonic motion) of a mass on a combined spring network.

Key relations: Series springs, $\dfrac{1}{k_{\text{eq}}}=\dfrac1{k_1}+\dfrac1{k_2}$; parallel springs, $k_{\text{eq}}=k_1+k_2$; $\omega_n=\sqrt{k_{\text{eff}}/m}$; SHM peaks $v_{\max}=\omega_n X$, $a_{\max}=\omega_n^2X$.

Why this works: The two left-hand springs share the same force but different deflections (series), while the reduced left-hand branch and the right-hand spring share the same displacement as the block moves (parallel) — reducing to one $k_{\text{eff}}$ turns the system into the standard $m\ddot x+k_{\text{eff}}x=0$ SHM equation whose solution's velocity and acceleration amplitudes follow directly from differentiating $x=X\sin(\omega_n t+\phi)$.

Common pitfall: Adding the series pair directly (treating them as parallel) or adding all three springs in parallel — always identify which springs share load (series) versus which share displacement (parallel) from the physical arrangement, not just their listed order.

Source: Hibbeler, Engineering Mechanics: Dynamics, Ch. 22 (Vibrations — Free Vibration of a Particle, Spring Combinations).

Question 6 of 6: Question 6 (paper Question VI)

Question 6 (paper Question VI) (20 marks)