PSL201Y1 Lecture Notes - Lecture 16: Central Pattern Generator, Cpg Site, Flat Feet

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10 Nov 2011

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Drive corresponding motor neurons on swing & Reflex loops operating on them especially. Stretch flexes, cutaneous reflexes (extensive thrust reinform activity in extensor burst gen while leg bearing weight) Propriospinal (in spinal) regulate reflex strength of swing & stance. Spinoreticulothalamic neurons have pain-related neurons near central canal so when very painful stimulus, tends to turn on some of these locomotor networks. So is foot stimulated, get increased flexion than normal, functional since means hit obstacle and increase flexion to get over it instead of stepping on hid. So dif connections in spinal cord, presynaptic inhibition, easy for ns to change signal effect. Swing when in air, not supporting weight: f starts w toe coming off ground (pure flexion, leg suspended in air) E1 when foot brought back to ground (extension, mostly passive gravity doing work so minimal muscle activity) Flexor burst generator: flexion phase of swing has a fixed duration, regardless of speed of locomotion ex.

Related Questions

Good Vibes: Introduction to Oscillations

The conditions that lead to simple harmonic motion are as follows:

There must be a position of stable equilibrium.

There must be a restoring force acting on the oscillating object. The direction of this force must always point toward the equilibrium, and its magnitude must be directly proportional to the magnitude of the object's displacement from its equilibrium position. Mathematically, the restoring force F⃗ is given by ⃗, where is the displacement from equilibrium and k is a constant that depends on the properties of the oscillating system.

The resistive forces in the system must be reasonably small.

Consider a block of mass m attached to a spring with force constant k, as shown in the figure(Figure 1) . The spring can be either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When the spring is relaxed, the block is located at x=0. If the block is pulled to the right a distance A and then released, A will be the amplitude of the resulting oscillations.

Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block.

Part A

After the block is released from x=A, it will

(a) remain at rest

(b) move to the left until it reaches equilibrium and stop there.

(d) move to the left until it reaches x=−A and then begin to move to the right.

The time it takes the block to complete one cycle is called the period. Usually, the period is denoted T and is measured in seconds.

The frequency, denoted f, is the number of cycles that are completed per unit of time: f=1/T. In SI units, f is measured in inverse seconds, or hertz (Hz).

Part B

If the period is doubled, the frequency is

(a) unchanged

(b) doubled

Part C

An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. What is its frequency f?

Express your answer in hertz.

Part D

If the frequency is 40 Hz, what is the period T?

Express your answer in seconds.

The following questions refer to the figure (Figure 2) that graphically depicts the oscillations of the block on the spring.

Note that the vertical axis represents the x-coordinate of the oscillating object, and the horizontal axis represents time.

Part E

Which points on the x-axis are located at a distance A from the equilibrium position?

(a) R only

(b) Q only

Part F

Suppose that the period is T. Which of the following points on the t-axis are separated by the time interval T?

(a) K and L

(b) K and M

(d) L and N

(e) M and P

Now assume for the remaining Parts G - J, that the x coordinate of point R is 0.12 m and the t coordinate of point K is 0.0050 s.

Part G

What is the period T?

Express your answer in seconds.

Part H

How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement?

Express your answer in seconds.

Part I

What distance d does the object cover during one period of oscillation?

Express your answer in meters.

Part J

What distance d does the object cover between the moments labeled K and N on the graph?

Express your answer in meters.

emeraldyak680

3. 3 Stratosphere-troposphere exchange

The rate of exchange of air between the troposphere and thestratosphere is critical for determining the potential of variouspollutants emitted from the surface to reach the stratosphere andaffect the stratospheric ozone layer. One of the first estimates ofthis rate was made in the 1960s using measurements of strontium-90(90Sr) in the stratosphere. Strontium-90 is a radioactive isotope(half-life 28 years) produced in nuclear explosions. It has nonatural sources. Large amounts of 90Sr were injected into thestratosphere in the 1950s by above-ground nuclear tests. Thesetests were banned by international treaty in 1962. Following thetest ban the stratospheric concentrations of 90Sr began to decreaseas 90Sr was transferred to the troposphere. In the troposphere,90Sr is removed by wet deposition with a lifetime of 10 days (bycontrast there is no rain, and hence no wet deposition, in thestratosphere). An intensive stratospheric measurement network wasoperated in the 1960s to monitor the decay of 90Sr in thestratosphere. We interpret these observations here using a 2-boxmodel for stratosphere-troposphere exchange with transfer rateconstants kTS and kST (yr-1) between the tropospheric andstratospheric reservoirs. The reservoirs are assumed to beindividually well-mixed.

Let mS and mT represent the masses of 90Sr in the stratosphere andin the troposphere respectively. Observations of the decrease inthe stratospheric inventory for the period 1963-1967 can be fittedto an exponential mS(t) = mS(0)exp(-kt) where k = 0.77 yr-1.

1. Write mass balance equations for mS and mT in the 1963-1967period.

2. Assuming that transfer of 90Sr from the troposphere to thestratosphere is negligible (we will verify this assumption later),show that the residence time of air in the stratosphere is tS =1/kST = 1.3 years.

3. Let m'T and m'S represent the total masses of air in thetroposphere and the stratosphere, respectively. Show that theresidence time of air in the troposphere is tT = tS (m'T/m'S) = 7.4years. Conclude as to the validity of your assumption in question2.

4. Hydrochlorofluorocarbons (HCFCs) have been adopted asreplacement products for the chlorofluorocarbons (CFCs), which werebanned by the Montreal protocol because of their harmful effect onthe ozone layer. In contrast to the CFCs, the HCFCs can be oxidizedin the troposphere, and the oxidation products washed out byprecipitation, so that most of the HCFCs do not penetrate into thestratosphere to destroy ozone. Two common HCFCs have trade namesHCFC-123 and HCFC-124; their lifetimes against oxidation in thetroposphere are 1.4 years and 5.9 years, respectively. There are noother sinks for these species in the troposphere. Using our 2-boxmodel, determine what fractions of the emitted HCFC-123 andHCFC-124 penetrate the stratosphere.

PSL201Y1 Lecture Notes - Lecture 16: Central Pattern Generator, Cpg Site, Flat Feet

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