Free Vibration of Single Degree Systems Questions and Answers

Answer: B

The natural frequency is \(\sqrt{\frac{5k}{2M}}\).

Take moments about O for a small rotation \(\theta\). The bar's centre is L/6 from O, the left spring is L/3 from O, and the right spring and the particle are 2L/3 from O.

\[ \begin{aligned} I_{O} &= \tfrac{ML^{2}}{12} + M\left(\tfrac{L}{6}\right)^{2} + \tfrac{M}{4}\left(\tfrac{2L}{3}\right)^{2} = \tfrac{2ML^{2}}{9} \\ k_{\theta} &= k\left(\tfrac{L}{3}\right)^{2} + k\left(\tfrac{2L}{3}\right)^{2} = \tfrac{5kL^{2}}{9} \end{aligned} \]

\[ \begin{aligned} \omega_{n} &= \sqrt{\frac{k_{\theta}}{I_{O}}} \\ &= \sqrt{\frac{5k}{2M}} \end{aligned} \]

Leaving out the particle's inertia gives \(\sqrt{5k/M}\); the 3k forms come from miscounting the spring arms.

Answer: D

The natural frequency is \(\sqrt{\frac{4k}{m}}\).

The rod is hinged at its right end, the mass sits at distance L from the hinge, and the spring acts at distance 2L. For a small rotation \(\theta\) about the hinge, the mass moves \(L\theta\) and the spring stretches \(2L\theta\). Taking moments about the hinge:

\[ \begin{aligned} mL^{2}\ddot{\theta} + k(2L)^{2}\theta &= 0 \\ \omega_{n}^{2} &= \frac{4kL^{2}}{mL^{2}} = \frac{4k}{m} \end{aligned} \]

\(\sqrt{k/m}\) treats the spring as if it were attached at the mass; \(\sqrt{2k/m}\) and \(\sqrt{k/2m}\) use the lever ratio 2 without squaring it, or invert it. The spring's lever arm enters squared because it affects both the spring force and that force's moment.

Answer: A

The natural frequency is \(\sqrt{\left(\frac{3EI}{L^{3}} + K\right)/m}\).

The mass hangs from the cantilever tip, and the spring also acts at the tip. A tip displacement \(x\) is resisted by the beam, with stiffness \(3EI/L^{3}\), and by the spring, with stiffness \(K\), at the same time. The two are in parallel, so their stiffnesses add:

\[ \begin{aligned} k_{eq} &= \frac{3EI}{L^{3}} + K \\ \omega_{n} &= \sqrt{k_{eq}/m} \end{aligned} \]

\(EI/L^{3}\) is not the tip stiffness of a cantilever (that is \(3EI/L^{3}\)), and \(2m\) has no basis since only one mass is present and the beam and spring are massless.

Answer: B

The natural frequency is \(\sqrt{k/m}\), unchanged by the rocket's acceleration.

In the rocket's frame the mass feels an effective gravity \(g + a\). That only shifts the static equilibrium position: the spring stretches by \(m(g + a)/k\) instead of \(mg/k\). Measuring displacement x from the new equilibrium, the equation of motion is still

\[ \begin{aligned} m\ddot{x} + kx &= 0 \\ \omega_{n} &= \sqrt{\frac{k}{m}} \end{aligned} \]

because a constant force does not change the restoring stiffness. The options with a reduced stiffness \(\bar{k}\), with \(g/a\), or with \(1 - a/g\) all wrongly let a constant acceleration alter the frequency. A pendulum's frequency would change, since its restoring force comes from gravity; a spring's does not.

Answer: D

The natural frequency is \(\sqrt{\dfrac{Ka^{2}}{Mb^{2}}}\).

Let the rod turn through a small angle \(\theta\) about B. The spring at A stretches by \(a\theta\) and the mass at C moves by \(b\theta\). Taking moments about B (gravity only sets the static equilibrium):

\[ \begin{aligned} Mb^{2}\ddot{\theta} + Ka^{2}\theta &= 0 \\ \omega_{n} &= \sqrt{\frac{Ka^{2}}{Mb^{2}}} \end{aligned} \]

In effect the spring acts at C with stiffness \(K(a/b)^{2}\). \(\sqrt{K/M}\) ignores the lever, \(\sqrt{Ka/(Mb)}\) uses the ratio only once, and \(\sqrt{Kb^{2}/(Ma^{2})}\) inverts it.

Answer: C

The natural frequency is \(\sqrt{\frac{3g}{2L}}\).

The rod is a compound pendulum. Its moment of inertia about the hinge at one end is \(mL^{2}/3\), and gravity acts at G, a distance \(L/2\) from the hinge. For small angles

\[ \begin{aligned} \frac{mL^{2}}{3}\ddot{\theta} + mg\frac{L}{2}\theta &= 0 \\ \omega_{n}^{2} &= \frac{mgL/2}{mL^{2}/3} = \frac{3g}{2L} \end{aligned} \]

\(\sqrt{g/L}\) treats the rod as a point mass at the tip, and \(\sqrt{2g/L}\) as a point mass at G; both ignore the rod's distributed inertia. \(\sqrt{6g/L}\) would follow from using the moment of inertia about G, \(mL^{2}/12\), instead of about the hinge.

Answer: D

The natural frequency is \(\sqrt{\frac{4k}{5m}}\).

The string runs from the cart over the fixed pulley, down around the movable pulley that carries the hanging mass, and up to a fixed point. When the cart moves x, the hanging mass moves x/2. With gravity only shifting the equilibrium, the energies are

\[ \begin{aligned} T &= \frac{1}{2}m\dot{x}^{2} + \frac{1}{2}m\left(\frac{\dot{x}}{2}\right)^{2} = \frac{1}{2}\cdot\frac{5m}{4}\dot{x}^{2} \\ V &= \frac{1}{2}kx^{2} \\ \omega_{n} &= \sqrt{\frac{k}{5m/4}} = \sqrt{\frac{4k}{5m}} \end{aligned} \]

\(\sqrt{k/(2m)}\) treats both masses as moving x, and \(\sqrt{4k/(3m)}\) and \(\sqrt{k/(3m)}\) do not correspond to the effective mass \(5m/4\).

Answer: A

The answer is 0.392 m and 1.26 s.

At static equilibrium the spring force balances the weight:

\[ \begin{aligned} \delta_{st} &= \frac{Mg}{k} = \frac{4 \times 9.81}{100} \\ &= 0.392\ \text{m} \end{aligned} \]

The natural period depends only on mass and stiffness:

\[ \begin{aligned} T &= 2\pi\sqrt{\frac{M}{k}} = 2\pi\sqrt{0.04} \\ &= 2\pi \times 0.2 = 1.26\ \text{s} \end{aligned} \]

(Equivalently \(T = 2\pi\sqrt{\delta_{st}/g}\), the same 1.26 s.) 0.626 m is \(\sqrt{0.392}\), a stray square root, and 3.52 s does not follow from \(2\pi\sqrt{M/k}\) with these values.

Answer: A

The natural frequency is \(\sqrt{\frac{4k}{M}}\).

If the mass moves down by x, the pulley moves down by x too, and both rope strands lengthen by x. The rope is inextensible, so the whole extra length 2x comes from the spring, whose tension rises by \(T = 2kx\). Both strands carry this tension, so the restoring force on the pulley and mass is

\[ \begin{aligned} F &= 2T = 4kx \\ k_{eq} &= 4k, \quad \omega_n = \sqrt{\frac{4k}{M}} \end{aligned} \]

\(\sqrt{2k/M}\) counts either the doubled spring extension or the two strands, but not both. \(\sqrt{k/M}\) treats the spring as if attached directly to the mass, and \(\sqrt{k/(2M)}\) inverts the mechanical advantage.

Answer: D

The motion is \(x(t) = P\cos\left(\sqrt{k/m}\,t\right)\).

An undamped spring-mass system obeys \(m\ddot{x} + kx = 0\), whose general solution is a sum of a cosine and a sine at the natural frequency \(\omega_{n} = \sqrt{k/m}\):

\[ \begin{aligned} x(t) &= A\cos\omega_{n}t + B\sin\omega_{n}t \\ x(0) &= A = P \\ \dot{x}(0) &= B\omega_{n} = 0 \Rightarrow B = 0 \end{aligned} \]

So only the cosine term survives, with amplitude P.

The sine-only answer starts at zero displacement, not at P. The sum \(P\cos + P\sin\) starts at P but with velocity \(P\omega_{n}\), not zero. The answer with P/2 on each term starts at only half the given displacement and also has a non-zero initial velocity.