Calculating Amplitude of SDOF System Under Impulse

This problem involves a single degree of freedom (SDOF) system that is initially at rest and is subjected to a short-duration impulsive force. The effect of an impulse on a system is to change its momentum, which in turn imparts a velocity to the system.

The given parameters for the SDOF system are:

• Mass, $\text{m} = 1 \text{ kg}$
• Stiffness, $\text{k} = 10 \text{ kN/m} = 10 \times 10^3 \text{ N/m} = 10000 \text{ N/m}$

The impulsive force has a magnitude of $5 \text{ kN}$ and acts for $10^{-4}$ seconds:

• Force magnitude, $\text{F} = 5 \text{ kN} = 5 \times 10^3 \text{ N} = 5000 \text{ N}$
• Duration, $\Delta \text{t} = 10^{-4} \text{ s}$

Step 1: Calculate the Impulse Magnitude

For a constant force acting over a short duration, the impulse is calculated as the product of the force magnitude and the duration.

$\text{Impulse} = \text{F} \times \Delta \text{t}$

Substituting the given values:

$\text{Impulse} = 5000 \text{ N} \times 10^{-4} \text{ s} = 0.5 \text{ Ns}$

Step 2: Determine the Initial Velocity Due to the Impulse

According to the impulse-momentum theorem, the impulse is equal to the change in momentum of the system. Since the system is initially at rest, the initial momentum is zero. The impulse imparts an initial velocity ($\text{v}_0$) to the mass.

$\text{Impulse} = \Delta (\text{momentum}) = \text{m} \times \text{v}_0 - \text{m} \times \text{v}_{\text{initial}}$

Since $\text{v}_{\text{initial}} = 0$:

$\text{Impulse} = \text{m} \times \text{v}_0$

We have the impulse (0.5 Ns) and the mass (1 kg), so we can find the initial velocity $\text{v}_0$ just after the impulse:

$0.5 \text{ Ns} = 1 \text{ kg} \times \text{v}_0$

$\text{v}_0 = \frac{0.5 \text{ Ns}}{1 \text{ kg}} = 0.5 \text{ m/s}$

Step 3: Calculate the Natural Frequency of the System

The natural frequency ($\omega_{\text{n}}$) of an undamped SDOF system is given by the formula:

$\omega_{\text{n}} = \sqrt{\frac{\text{k}}{\text{m}}}$

Substituting the values of stiffness and mass:

$\omega_{\text{n}} = \sqrt{\frac{10000 \text{ N/m}}{1 \text{ kg}}} = \sqrt{10000} \text{ rad/s} = 100 \text{ rad/s}$

Step 4: Determine the Amplitude of Resulting Free Vibration

The system starts vibrating freely after the impulse. Since the system was initially at rest, the initial displacement $\text{x}(0)$ is 0. The impulse imparted an initial velocity $\text{v}_0$ to the system. For an undamped free vibration with $\text{x}(0)=0$ and $\text{v}(0)=\text{v}_0$, the displacement is given by $\text{x(t)} = \text{A} \sin(\omega_{\text{n}} \text{t})$, where A is the amplitude. The amplitude A is given by:

$\text{A} = \frac{\text{v}_0}{\omega_{\text{n}}}$

Substituting the calculated initial velocity and natural frequency:

$\text{A} = \frac{0.5 \text{ m/s}}{100 \text{ rad/s}} = 0.005 \text{ m}$

Step 5: Convert Amplitude to Millimeters

The question asks for the amplitude in millimeters (mm). We need to convert the amplitude from meters to millimeters.

$1 \text{ m} = 1000 \text{ mm}$

$\text{A}_{\text{(mm)}} = \text{A}_{\text{(m)}} \times 1000 \text{ mm/m}$

$\text{A}_{\text{(mm)}} = 0.005 \text{ m} \times 1000 \text{ mm/m} = 5.0 \text{ mm}$

The amplitude of the resulting free vibration is 5.0 mm.

Let's check the given options:

• 0.5 mm
• 10.0 mm
• 1.0 mm
• 5.0 mm

Our calculated amplitude is 5.0 mm, which matches one of the options.