A) A spring has an unstretched length of 10 cm. A 100-g mass...

a) A spring has an unstretched length of 10 cm. A 100-g mass is added, and the spring stretches to a total length of 20 cm, where the mass rests at equilibrium. What is the spring constant? b) Now, the same mass as in problem 1 is pulled down an additional 5 cm, so that the length of the spring is 25 cm, and then released. Draw a graph that shows the subsequent motion of the spring (as a function of time). On your graph, show the meaning of the amplitude and the period—include numbers! I need help with question b. The answer to question a is 9.8 N/m.

Transcript

00:01 Oh, here we have to solve the following problem.

00:02 Let's start with a.

00:04 Here, spring has a stretch length of 10 centimeters, and then 100 grams weight is added, and the spring stretch is to the total length of 20 centimeters.

00:22 First, we have to calculate the spring constant.

00:26 The spring constant equals to mg.

00:30 Divided by delta l which is m g divided by l final minus l0 so that is 0 .100 kilograms times 9 .80 meters per second squared divided by 0 .10 meters that is 9 .8 newton per meter and now let's answer question b so now now it's pulled to an additional pull down to an additional 5 centimeters.

01:12 So therefore the amplitude is 5 centimeters.

01:18 And then released.

01:21 And now we have to draw the graph which shows the motion of the spring as a function of time.

01:28 And we have to indicate the amplitude and the period.

01:31 So let's do this.

01:33 Let's illustrate it so after the weight was added and the spring was deformed to 20 centimeters it was pulled down further and deformed to 25 centimeters therefore that's an amplitude of the motion which is a and here let's introduce y -axis upwards so therefore the initial so y, therefore y as a function of time equals to negative a times cosine of omega t.

02:14 Where omega, where that is negative 0 .05 meters times, or actually we can write down as 5 centimeters, negative 5 .0 .0 centimeters times, comes cosine of omega and here omega equals to square root of k over m which is 9 .9 which is 9 .90 newton sorry radiance per second.

03:23 Therefore the period equals to 2 pi over omega...

Question

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