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This is one of our many quizzes on Sinusoidal graph properties. Click on this to access more such competitive quizzes on Trigonometry.

Sinusoidal graph properties Quiz 2

mathematics

 Which of the following sinusoidal waves has amplitude equal to $6$ and period equal to $\dfrac{\pi }{2}$?

 Use the formula for finding the amplitude and period of the wave $y=A\cos (Bx+C)+D$, and eliminate the wrong answers.

  For the sinusoidal waves $y=A\sin (Bx+C)+D$ and $y=A\cos (Bx+C)+D$, amplitude of sinusoidal wave is $\left| A \right|$ and period of the wave is equal to $\dfrac{2\pi }{\left| B \right|}$. $ \\ $ For wave $y=6\cos \left( 4x \right)$, $ \\ $ Amplitude is $6$ and the period is $\dfrac{2\pi }{4}=\dfrac{\pi }{2}$. $ \\ $ For wave $y=6\cos \left( 2x \right)$, $ \\ $ Amplitude is $6$ and period is $\dfrac{2\pi }{2}=\pi $. $ \\ $ For wave $y=3\cos \left( 4x \right)+2$, $ \\ $ Amplitude is $3$ and period is $\dfrac{2\pi }{4}=\dfrac{\pi }{2}$. $ \\ $ For wave $y=6\cos \left( 2x \right)+2$, $ \\ $ Amplitude is $6$ and period is $\dfrac{2\pi }{2}=\pi $.

 For how many set(s) of values of $x\text{ and }y$; $0\le \theta \le {{90}^{{}^\circ }}$, do the waves ${{y}_{1}}=6\sin {{\left( 2x-4y \right)}^{{}^\circ }}\text{ and }{{y}_{2}}=6\cos {{\left( 4x-2y \right)}^{{}^\circ }}$ give the same value?

 Assume $\theta ={{45}^{{}^\circ }}$ as, for $0\le \theta \le {{90}^{{}^\circ }}$, sinusoidal waves $y=A\sin \theta $ and $y=A\cos \theta $ give same value at $\theta ={{45}^{{}^\circ }}$ and solve for the sets of values of $x\text{ and }y$.

  For $0\le \theta \le {{90}^{{}^\circ }}$, sinusoidal waves $y=A\sin \theta $ and $y=A\cos \theta $ give same value at$\theta ={{45}^{{}^\circ }}$. $\begin{align}  & \Rightarrow 2x-3y={{45}^{{}^\circ }}\text{ and 4}x-2y={{45}^{{}^\circ }} \\  & \Rightarrow 2x-3y=\text{4}x-2y \\  & \Rightarrow 2x=-y \\  & \Rightarrow x=\dfrac{-y}{2} \\ \end{align}$Thus, for different values of $x$ we have distinct values of $y.$ $ \\ $ So, there are infinite sets of $\left( x,y \right)$ which satisfy $6\sin {{\left( 2x-4y \right)}^{{}^\circ }}=6\cos {{\left( 4x-2y \right)}^{{}^\circ }}$.

 If the period of the cosine wave $y=4\cos \left( ax-5 \right)$ is equal to $\dfrac{\pi }{5}$ and the period of the sine wave $y=8\sin \left( 4bx-5 \right)+6$ is equal to $\dfrac{\pi }{10}$, then period of cosine wave $y=10\cos \left[ \left( {{a}^{2}}+{{b}^{2}} \right)\pi x-5 \right]+21$ is

 Use the formula to find the period of the sinusoidal wave and determine the values of $a$ and $b$, and further evaluate the value of ${{a}^{2}}+{{b}^{2}}$.

Period of $y=A\cos (Bx-C)+D$ and $y=A\sin (Bx-C)+D$ is equal to $\dfrac{2\pi }{\left| B \right|}$. $ \\ $ Thus, the period of the cosine wave $y=4\cos \left( ax-5 \right)$ is $\dfrac{\pi }{5}$. $\begin{align}  & \Rightarrow \dfrac{2\pi }{\left| a \right|}=\dfrac{\pi }{5} \\   & \Rightarrow a=10 \\ \end{align}$ And, the period of the sine wave $y=8\sin \left( 4bx-5 \right)+6$ is equal to $\dfrac{\pi }{10}$. $\begin{align}  & \Rightarrow \dfrac{2\pi }{\left| 4b \right|}=\dfrac{\pi }{10} \\  & \Rightarrow b=5 \\ \end{align}$So, ${{a}^{2}}+{{b}^{2}}$is equal to,$\begin{align}  & \Rightarrow {{\left( 10 \right)}^{2}}+{{\left( 5 \right)}^{2}} \\  & \Rightarrow 125 \\ \end{align}$So, period of cosine wave $y=10\cos \left[ \left( {{a}^{2}}+{{b}^{2}} \right)\pi x-5 \right]+21$ is, $ \\ $ $\Rightarrow \dfrac{2\pi }{\left| 125\pi  \right|}=\dfrac{2}{125}$

 How many sine waves with equation $y=a\sin \left[ \left( 4{{a}^{2}}-1 \right)x-2 \right]+2$ have a period equal to $\pi $?

 Use the formula of the period of sine wave and obtain the value(s) of ‘a’. For, different values of ‘a’, obtain different sine waves.

 The period of $y=A\sin (Bx-C)+D$ is equal to $\dfrac{2\pi }{\left| B \right|}$. $ \\ $ So, for function $y=a\sin \left[ \left( 4{{a}^{2}}-1 \right)x-2 \right]+2$ with period $\pi $, $\begin{align}   & \Rightarrow \dfrac{2\pi }{4{{a}^{2}}-1}=\pi  \\  & \Rightarrow a=\pm \dfrac{\sqrt{3}}{2} \\ \end{align}$ Thus, the functions are $y=\dfrac{\sqrt{3}}{2}\sin \left( 2x-2 \right)+2$ and $y=-\dfrac{\sqrt{3}}{2}\sin \left( 2x-2 \right)+2$.

 On graphing paper, the path followed by a white knuckle ride from point A to C is plotted whose equation is given by sine wave $y=2\sin \left( \dfrac{x}{10}-1.55 \right)+D$. Let, P and Q are two ticket counters to buy tickets for the ride, and line PQ is the midline of the sine wave of the ride on which point B lies. If AC is ground and the area of triangle with base $64\text{ m}$ is equal to $1280\text{ }{{\text{m}}^{2}}$, then by considering the figure given below answer the following questions $ \\ $ 1. What is the distance between points P and Q? $ \\ $ 2. What is the maximum height ‘D’ attained by the ride? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e443e50e-0bd3-446a-9742-021d168f94e02523493154599241417.png"/>

 Visualize the figure carefully and find the period of sine wave to evaluate length PQ. Use the formula of triangle and concept of midline to evaluate the value of D.

 From the figure it can be clearly seen that from P and Q, the roller coaster passes and goes upward and as PQ is the midline of the sine wave. $ \\ $ Thus, PQ is equal to the period of the sine wave $y=2\sin \left( \dfrac{x}{10}-1.55 \right)+D$. $ \\ $ Period of sine wave $y=A\sin \left( Bx-C \right)+D$ is $\dfrac{2\pi }{\left| B \right|}$. $\begin{align}  & \Rightarrow \text{PQ}=\dfrac{2\pi }{\left| \dfrac{1}{10} \right|}\text{ m} \\  & \Rightarrow \text{PQ}=20\pi \text{ m} \\ \end{align}$ Area of the triangle with base ‘b’ and height ‘h’ is $\dfrac{bh}{2}$. $ \\ $ In triangle ABC, $\begin{align}  & \Rightarrow \dfrac{AC\times BO}{2}=1280\text{ }{{\text{m}}^{2}} \\  & \Rightarrow \dfrac{64\times BO}{2}=1280\text{ }{{\text{m}}^{2}} \\  & \Rightarrow BO=40\text{ m} \\ \end{align}$ As, PQ is midline of the sine wave $y=2\sin \left( \dfrac{x}{10}-1.55 \right)+D$, $ \\ $ Then, from figure $\begin{align}  & \Rightarrow D=2\left( 40 \right)\text{ m} \\  & \Rightarrow D=80\text{ m} \\ \end{align}$