A particle is executing SHM along a straight line. Its velocities at distance x1 and x2 are v1 and v2, respectively. What is its time period?
If A is amplitude of S.H.M. and ω is angular frequency,the velocity as function of displacement x is given by[math]v=��\sqrt{(A^2-x^2)}[/math].Squaring on both sides,we have, [math]v^2=ω^2(A^2-x^2)[/math] .Then, [math]v_1^2=ω^2(A^2-x_1 ^2)[/math]…………………..(1).Similarly, [math]v_2^2=ω^2(A^2-x_2^2)[/math]………………….(2)Subtracting (1) from(2), we get,[math]v_2^2-v_1^2=ω^2(x_1^2-x_2^2)[/math]. Taking ω[math]=2pi/T [/math]and making T subject of the formula,we finally get[math]T=\frac{2π\sqrt{(x_1^2-x_2^2)}}{\sqrt{(v_2^2-v_1^2)}} [/math].
Match each interval with its corresponding average rate of change for the graph shown below (see pic)?
Match please (see attached graph) 1. 0 ≤ x ≤ 1 a) -0.5 2. -2 ≤ x ≤ -1 b) -1.5 3. -2 ≤ x ≤ 0 c) -0.375 4. 1 ≤ x ≤ 2 d) -0.25 5. -1 ≤ x ≤ 0 e) -2 6. 0 ≤ x ≤ 2 f) -1
How do I estimate the standard deviation off of a Histogram?
Normally construction of a histogram is preceded by preparation of frequency distribution table. If you have only a histogram, doesn't matter. Do the reverse job, make a rather detailed frequency distribution table having the following columns:(1) class interval(2) mid value of class interval(3) frequency(4) mid value x frequency(5) mid value - meanFinding out Mean from histogram (or grouped data)Find out the total of (mid value x frequency). Divide it by number of observations. You get mean.Finding out Standard Deviation from histogram (or grouped data)Make a separate column for the purpose of calculating standard deviation. In this column, write down the value of the following term against each class interval.frequency x square of (mid value -mean)Now find out the total of this column. Divide it by number of observations and take square root. That is standard deviation.
A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, what is its displacement y?
We know that velocity of particle at displacement x is given byv=w(A^2-x^2)^1/2.A is amplitude and w is angular frequency of SHM.The maximum velocity is Aw. So, when v=Aw/2, we haveAw/2=w(A^2-x^2)^1/2.Squaring on both sides,A^2w^2/4=w^2A^2-w^2x^2 ORx^2= (3/4)A^2 ORx= (3/4)^1/2 A.