An undamped oscillator has period ro = 1.000 s, but i now add a little damping so that its period changes to r i = 1.001 s. what is the damping factor 8? by what factor will the amplitude of oscillation decrease after 10 cycles? which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?

Step-by-step explanation:

a) Under the undamped oscillator, the equation that relates frequencies is:

           (1)

when β is the damping factor and ω₀ is the natural frequency.

Let's solve the equation (1) for β.

.

So  β = 0.2808 1/s.

b) Now, the amplitud equation is: , after 10 cycles we will have t=10*T₁ and .

In says that after 10 cycles the amplitude of oscillation decrease by a factor of 0.06 or about 6%. In the case of the period, we have 0.1%, concluding that the change of amplitude is more noticeable.  

Have a nice day!

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.

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