Download presentation

Presentation is loading. Please wait.

1
**Horizontal Spring-Block Oscillators**

A popular demonstration vehicle for simple harmonic motion is the spring block oscillator.

2
**Oscillations of a Spring**

Displacement is measured from the equilibrium point. Amplitude is the maximum displacement. A cycle is a full to-and-from motion. Period is the time required to complete one cycle. Frequency is the number of cycles completed per second. Figure Caption: Force on, and velocity of, a mass at different positions of its oscillation cycle on a frictionless surface.

3
**The block-spring system shown on the right forms a linear SHM oscillator.**

The spring constant of the spring, k, is related to the angular frequency, w, of the oscillator:

4
Spring Problem A 1.0 kg mass attached to one end of a spring completes one oscillation every 2.0 s. Find the spring constant.

5
Do Now 1. A block whose mass m is 5kg is fastened to a spring whose spring constant k is 2000 N/m (See figure below). The block is pulled a distance x = 8 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0 Determine the period of oscillation, the frequency, and the angular frequency for the block.

6
**Ts = 2π 𝑚 𝑘 = 2π 5𝑘𝑔 2000𝑁/𝑚 = 0. 314s f = 1 𝑇 = 1 0. 314𝑠 = 3**

Ts = 2π 𝑚 𝑘 = 2π 5𝑘𝑔 2000𝑁/𝑚 = 0.314s f = 1 𝑇 = 𝑠 = 3.18 Hz ω = 2πf =2π(3.18Hz) = 20 rad/s

7
X - X Displacement (x) Velocity (v) max Potential energy (U) Kinetic energy (K) Force (F) - max Acceleration (a)

8
Do Now

14
Note: In a vertical motion, the spring is not in equilibrium at its unstrained length, but at the point where the hanging block is in equilibrium.

15
The restoring force also leads to simple harmonic motion when the object is attached to a vertical spring, just as it does when the spring is horizontal. When the spring is vertical, however, the weight of the object causes the spring to stretch, and the motion occurs with respect to the equilibrium position of the object on the stretched spring . mg = kd0, which gives d0 = mg/k.

18
At equilibrium, Fnet =0 Fnet = mg – kyeq = 0 yeq = 𝑚𝑔 𝑘 When the mass is pulled A displacement Fnet = mg – k(yeq + A) = mg – kyeq – kA = 0 – kA = - kA

19
A 2-kg block attached to an un-stretched spring of spring constant k=200N/m as shown in the diagram below is released from rest. Determine the period of the block’s oscillation.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google