# Tutor profile: Vignesh K.

## Questions

### Subject: Trigonometry

You want to find out the height of a tall building, but all you have is a 50 m tape. . How will you figure out the approximate height of the structure ? Assume suitable values to illustrate an example of your solution.

The building can be assumed as a straight line going vertically up , perpendicular to the ground. Let the meeting point and the ground be 'O' and the topmost point be 'H', the distance between O and H is height of the building. Let it be denoted as 'h' One solution to find the height of the building using trigonometry: Fully extend the tape and place one end at 'O' and the other end at 'C'. Be careful to place the tape perpendicular to the building (on the ground) Now take a picture with the tape and the building in the frame. Mark the corresponding points O, C and H , then join the points O and H to get line OH which represents the building and join points O and C which represents the tape, and finally join CH. Now, you obtain a right angle triangle COH , using a protractor measure the angle HCO and denote it a '$$\theta$$'. Use the following trigonometric identity; $$tan(\theta) = $$ (length of opposite side) / (length of adjacent side); which gives $$tan(\theta)$$ = OH/OC using this Height of the building ; OH = h =$$ tan(\theta)$$ * OC Example for illustration: Suppose I measured the angle $$\theta $$= 60 degree. h = tan(60) * OC we know OC = 50 m; Therefore; h = $$sqrt(3)$$ * 50; h = 86.6 m Therefore the approximate height of the building is 87m.

### Subject: Geometry

During the assembly of my machine I need to connect two pulleys with a belt. The pulleys are situated on the horizontal axis and have their centers co-linear. They have an equal diameter of 5 cm. the total length of the belt is 1 m. 1. What must be the center to center distance between the two pulleys to keep the belt tight? 2. Suppose I change the Diameter of one of the pulleys to 10 cm ; what must be the new center to center distance between the pulleys?

Statement 1 : For the belt to be held tightly in position the belt must go around half the circumference of each of the pulley. ( Pulleys are assumed as circles ) $$\textbf{Case 1} : $$ Let ; total length of the pulley; 'l' = 1000 mm; Diameter of each of the pulley 'd' = 50 mm; let the unknown ; center to center distance be 'x' So according to the statement 1 total length = 2 * center to center distance + 2 * (half the circumference of one pulley) l = 2*x + 2*$$\pi$$ *r ; formula used : circumference of a circle = 2*pi*r ( r = radius of the circle ; r = d/2) On substitution of the values; 1000 = 2*x + 2 * 3.14 * 25 2*x = 1000 - 2*3.14*25 2*x = 843; x = 421.5 mm Therefore the center distance between the pulleys must be approximately 422 mm $$\textbf{Case 2} : $$ In the first case since the pulleys had equal diameters the belt remains parallel to the horizontal axis. But when diameter of one of the pulleys is increased, the belt is inclined to the horizontal axis. Let 'R' be radius of big pulley and 'r' be radius of small pulley. Vertical lines are drawn form center of each of the pulleys to intersect the small and the big pulley at 'A' and 'B' respectively. Draw a line joining A and B and let the distance between A and B be 'y' r = 25 mm and R = 50 mm; So according to statement 1 l = 2*y+ $$\pi$$ *r + $$\pi$$ *R; On substitution of the values; 1000 = 2*y + 3.14 * 25 + 3.14 *50; 2y = 1000 - 3.14 * 25 + 3.14 *50 ; 2y = 764.5; y = 382.25 mm y is the inclined distance; but what is asked is the horizontal distance. From the point A draw a horizontal line to intersect the vertical line drawn earlier of big pulley at C. ABC forms a right angle triangle where; AC = 'x' = center to center distance; AB= 'y' = 382.25 mm ; BC = R-r = 25 mm; According to $$\textit{Pythagoras' theorem}$$; $$ AB^{2} = AC^{2} + BC^{2} $$ On substituting the values $$382.25^{2} = x^{2} +25^{2};$$ $$x^{2} = 145490.0625;$$ x = 381.43 mm Therefore for Case 2 ; the approximate center distance between the pulleys is 381 mm $$\textit{It's recommended to draw a figure for a better understanding of the problem}$$

### Subject: Differential Equations

The governing Differential Equation for a Simple Pendulum is $$ \ddot{\theta} + \frac{g}{l} \theta = 0$$ Solve the above differential equation for the Initial conditions: $$\theta_{0} = 5 $$ ; $$\dot{\theta} = 0$$ Also give a physical explanation of the solution obtained Units for the above problem ; length in meters, angle ( $$\theta $$ ) in degree and time in seconds. Assume 'g' ( acceleration due to gravity ) = 9.81 m/$$ s^{2}$$ ; $$\omega $$ for the pendulum = 1 deg/sec Assumptions used for deriving the differential equation 1. Oscillations are small 2. String used is mass less and in-extensible

The given equations is a linear second order differential equation. To solve the equation we first assume a solution; Let, $$ \theta = e^{mt} $$ ( where m is an constant and t is the independent variable of the equation ) On substitution into the equation we get ; $$ ( m^{2} + \frac{g}{l} ) * (e^{mt}) = 0$$ ( formula used $$\ddot{\theta} = m^{2} * e^{mt} $$ ) Since $$e^{mt} \neq 0 $$ , which is a trivial solution; $$ ( m^{2} + \frac{g}{l} ) = 0$$ ; $$m = \pm \sqrt{\frac{-g}{l}}$$; or $$m = i \omega; $$ where $$ i = \sqrt{-1} $$ and $$\omega = \sqrt{\frac{g}{l}}$$; Therefore the $$\textbf{General Solution} $$ of the differential equation is; $$ \theta = A e^{i\omega} + Be^{-i\omega}$$ or $$ \theta = A sin(\omega t) + Bcos (\omega t)$$ To get the particular solution we make use of the boundary condition given in the problem; On substitution of the first boundary condition to the general solution; 5 = Asin(0) + Bcos(0); (given $$\omega = 1$$) B = 5; On substitution of the second boundary condition to the general solution; $$0 = A\omega cos(0) - B\omega sin(0) $$ A =0; Therefore he $$\textbf{Particular Solution}$$ for the differential equation is $$\theta = 5 cos(t) $$ Physical Interpretation of the solution : The boundary conditions imply the that simple pendulum is displaced by 5 degree (1st boundary condition) and then released without imparting any velocity to it (2nd boundary condition) The solution implies that in the absence of losses the pendulum will keep oscillation about its mean position with an amplitude 5 degree and frequency of 1 deg/sec. This motion is simple harmonic motion

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