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1.To recognise that a loaded spring is an application of SHM 2.To review Hooke’s law 3.To establish the period of an oscillating loaded spring 4.To review.

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Presentation on theme: "1.To recognise that a loaded spring is an application of SHM 2.To review Hooke’s law 3.To establish the period of an oscillating loaded spring 4.To review."— Presentation transcript:

1 1.To recognise that a loaded spring is an application of SHM 2.To review Hooke’s law 3.To establish the period of an oscillating loaded spring 4.To review springs in parallel & series Book Reference : Pages 41-42

2 What can we remember from Y12? What does Hooke’s law state? How do we express this as a proportionality? How do we turn this into an equation we can use?

3 http://www.rdg.ac.uk/acadepts/sp/picetl/publish/ISEs/Forces2.htm Hooke’s Law states that...the change in length produced by a force on a wire or spring is directly proportional to the force applied.  L (m) Hooke’s law only applies within limits Extension  Force Applied  L  F

4 To turn a proportionality into an equation introduce a constant of proportionality...  L  F F = k  L We call k the spring constant and it defines how stiff the spring is How can we find K experimentally?

5 As a mass m on a spring oscillates, the increase in spring tension (  T s ) provides the “restoring force” required for SHM At displacement x from the equilibrium position we can calculate the restoring force from Hooke’s law  T s = -kx (minus sign shows force acting towards the equilibrium position)

6 So the restoring force is –kx and we can relate this to acceleration via (f=ma) (a=f/m) Acceleration = restoring force / mass a = -kx/m For SHM we also know that the acceleration can be written as a = -(2  f) 2 x

7 Equating the two views of a -(2  f) 2 x = -kx/m (then cancel x & times by -1 to kill - sign) (2  f) 2 = k/m (then square root) 2  f =  (k/m) (then ÷ 2  ) f =  (k/m) / 2  (find period T by knowing it is 1/f) (÷ both sides by f, times both sides by 2 , ÷ throughout by  (k/m) (actually we times by inverted devisor  (m/k) 1/f = 2   (m/k) = T (find this in the crib sheet & highlight)

8 When will the tension in the spring be at a maximum? What will the maximum tension be equal to? When will the tension in the spring be at a minimum? What will the minimum tension be equal to? Equilibrium position x = -A x = +A

9 Tension varies between mg + kA and mg - kA At a maximum when spring is fully stretched at -A At a minimum when spring is stretched as little as possible at +A Equilibrium position x = -A x = +A

10 What happens to the frequency when the mass increases? What happens to the frequency when the spring constant increases, (stiffer spring)? How would moving the experiment to the moon change the period or frequency?

11 f =  (k/m) / 2  If the mass increases the frequency decreases, more inertia The frequency increases if the spring constant increases The equation for frequency/period of a loaded spring does not depend upon g. Therefore moving it to the moon will change nothing (unlike the simple pendulum)

12 Thinking about this experimentally, and compare with the simple pendulum experiment What could we change? What could we measure? What could we plot? What could we find from the graph?

13 The mass m can be our independent variable We can measure the period by finding the time for n full oscillations and dividing appropriately We want a straight line graph... T 2 against m, (usual dependent vs independent variable) Gradient will be 4  2 /k What should we do to ensure the best quality value for this or any gradient?

14 Springs in parallel share the load & have the same extension acting like a single spring with a combined spring constant Derive an equation for the combined spring constant in terms of the two spring constants k p & k q pq LL Hints : identical extensions, different tensions, resolve vertically

15 The force needed to stretch springs p & q respectively is:- F p = k p  L & F q = k q  L W = k p  L + k q  L which can be considered equal to k  L where k is the effective spring constant k  L = k p  L + k q  L remove  L k = k p + k q pq LL The tension is given by W = F p + F q and so...

16 Springs in series share the same tension which is equal to W Derive an equation for the combined spring constant in terms of the two spring constants k p & k q p q LL Hints : different extensions, identical tensions, equate the total extension to the sum of the two extensions

17 Springs in series share the same tension which is equal to W The extensions in the two springs is given by:-  L p = W/k p &  L q = W/k q The total extension is  L p +  L q = W/k p + W/k q = W/k = 1/k p + 1/k q = 1/k where k is the effective spring constant p q LL


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