A 9.0 kg mass is attached to a spring and oscillates on a horizontal frictionless surface. A 2.0 kg mass sits on top of the 9.0 kg as it oscillates. The spring constant of the spring causing the motion is 80 n/m
The correct answer and explanation is:
Let’s analyze the problem step-by-step:
Given Data:
- Mass 1, m1=9.0 kgm_1 = 9.0 \, \text{kg} (attached to the spring)
- Mass 2, m2=2.0 kgm_2 = 2.0 \, \text{kg} (sitting on top of mass 1)
- Spring constant, k=80 N/mk = 80 \, \text{N/m}
- Surface is frictionless, so no energy lost due to friction
What is being asked?
While you didn’t specify an explicit question, usually in such a problem, typical questions could be:
- What is the period of oscillation?
- What is the frequency of oscillation?
- What is the angular frequency?
- How does the presence of the 2.0 kg mass affect the system?
Step 1: Understand the System
Both masses m1m_1 and m2m_2 are oscillating together because the 2.0 kg mass sits on top of the 9.0 kg mass. The system moves as one unit horizontally (no friction means the top mass won’t slip).
So, the total effective mass oscillating is: mtotal=m1+m2=9.0 kg+2.0 kg=11.0 kgm_{\text{total}} = m_1 + m_2 = 9.0 \, \text{kg} + 2.0 \, \text{kg} = 11.0 \, \text{kg}
Step 2: Calculate Angular Frequency ω\omega
For a mass-spring system, the angular frequency ω\omega is: ω=kmtotal\omega = \sqrt{\frac{k}{m_{\text{total}}}}
Substitute values: ω=80 N/m11.0 kg=7.2727≈2.698 rad/s\omega = \sqrt{\frac{80 \, \text{N/m}}{11.0 \, \text{kg}}} = \sqrt{7.2727} \approx 2.698 \, \text{rad/s}
Step 3: Calculate the Period TT
The period TT is the time for one complete oscillation: T=2πωT = \frac{2\pi}{\omega}
Calculate: T=2π2.698≈6.2832.698≈2.33 secondsT = \frac{2\pi}{2.698} \approx \frac{6.283}{2.698} \approx 2.33 \, \text{seconds}
Step 4: Calculate the Frequency ff
Frequency is the reciprocal of the period: f=1T=ω2π=2.6986.283≈0.429 Hzf = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{2.698}{6.283} \approx 0.429 \, \text{Hz}
Summary of Key Results:
- Total mass oscillating: 11.0 kg
- Angular frequency, ω≈2.70 rad/s\omega \approx 2.70 \, \text{rad/s}
- Period, T≈2.33 secondsT \approx 2.33 \, \text{seconds}
- Frequency, f≈0.43 Hzf \approx 0.43 \, \text{Hz}
Explanation:
In a horizontal spring-mass system without friction, the period of oscillation depends only on the spring constant and the total mass attached. Since the 2.0 kg mass sits on top of the 9.0 kg mass and moves with it, their masses add up to give a combined oscillating mass of 11.0 kg.
The spring constant kk represents the stiffness of the spring — the higher the kk, the stronger the restoring force, and thus faster oscillations (shorter period). Meanwhile, the more mass attached, the slower the oscillations because it takes more time to move a heavier mass.
Frictionless surfaces ensure energy is conserved in the system, so the oscillations continue without damping. This idealized scenario lets us use the simple harmonic motion formulas directly.
This problem highlights how additional mass affects oscillation: adding more mass lowers the frequency and increases the period. For practical applications, such as designing suspension systems or vibration absorbers, knowing how mass affects oscillations is crucial.