📝 Sine and cosine functions amplitude period (16 MCQs)
📖 From Calculus • 1. Basics before calculus • 16 questions available
What is Sine and cosine functions amplitude period?
Definition:
The sine and cosine functions and have amplitude (half the vertical distance between max and min) and period (horizontal length for one complete cycle).
Example:
For , amplitude = 3, period = .
Reason:
Amplitude and period describe wave characteristics, essential in physics (sound, light), engineering (signals), and any periodic phenomenon.
📝 All Sine and cosine functions amplitude period MCQs
Q1. If the period of \y = A\\sin Bx\ is halved, what happens to the constant \B\?
📖 Explanation: Doubling the period means the wavelength is cut in half, so the angular frequency \B\ must increase to keep the relationship \\\text{period}=\\frac{2\\pi}{B}\ valid. Therefore, when the period is halved, \B\ becomes twice its original value, i.e., it doubles.
Q2. Given \y = 3\\sin(2x)\ and \y = 4\\cos(2x)\, which function has the larger absolute value at \x = \\frac{\\pi}{4}\?
📖 Explanation: At \x = \\frac{\\pi}{4}\, compute \2x = \\frac{\\pi}{2}\. Then \\\sin(\\frac{\\pi}{2}) = 1\ giving \y = 3\. Meanwhile \\\cos(\\frac{\\pi}{2}) = 0\ giving \y = 0\. Hence the sine function yields an absolute value of 3, which is larger than the cosine function's value of 0.
Q3. Two functions \y_1 = A\\sin(Bx)\ and \y_2 = A\\cos(Bx)\ intersect at \x = 0\. What is the smallest positive \x\ where they intersect again?
📖 Explanation: Intersection requires \\\sin(Bx) = \\cos(Bx)\, which implies \\\tan(Bx) = 1\. The general solution is \Bx = \\frac{\\pi}{4} + n\\pi\. The smallest positive solution occurs when \n = 0\, giving \Bx = \\frac{\\pi}{4}\ and thus \x = \\frac{\\pi}{4B}\.
Q4. Reflecting the graph of \y = A\\sin Bx\ over the x‑axis results in which expression?
📖 Explanation: Reflecting over the x‑axis changes the sign of every y‑coordinate while leaving the x‑coordinates unchanged. Algebraically this corresponds to multiplying the entire function by \-1\. Hence the reflected function is \-A\\sin(Bx)\. The other options either keep the sign positive or introduce an unnecessary sign change inside the argument.
Q5. For \y = A\\sin Bx\, the total area under one period of \|y|\ is approximated by which expression?
📖 Explanation: The absolute value \|\\sin\\theta|\ over a full period integrates to 4. Scaling by \A\ vertically and compressing horizontally by \B\ changes the integral to \\\frac{4A}{B}\. This result follows from \\\int_{0}^{2\\pi}|\\sin\\theta|\\,d\\theta = 4\ and the substitution \\\theta = Bx\.
Q6. Which statement correctly describes the phase relationship between \y = A\\sin Bx\ and \y = A\\cos Bx\?
📖 Explanation: The cosine function can be written as \\\cos(Bx) = \\sin(Bx + \\frac{\\pi}{2})\. This shows that the cosine wave reaches the same points as the sine wave, but \\\frac{\\pi}{2}\ radians earlier, meaning cosine leads sine by a quarter‑cycle (\\\frac{\\pi}{2}\).
Q7. What is the average value of \y = A\\sin Bx\ over one full period compared to \y = A\\cos Bx\?
📖 Explanation: Both sine and cosine are symmetric about the horizontal axis. Over a complete period, the positive and negative contributions cancel out, yielding an average (mean) value of zero for each function regardless of the amplitude \A\ or frequency \B\. Hence their averages are equal.
Q8. Increasing \A\ versus increasing \B\ affects the graph how?
📖 Explanation: The parameter \A\ multiplies the function value, so larger \A\ stretches the graph away from the x‑axis (vertical stretch). The parameter \B\ appears inside the argument; larger \B\ reduces the period \\\frac{2\\pi}{B}\, causing the wave to repeat more quickly, i.e., a horizontal compression.
Q9. Which function has the larger maximum rate of change: \y_1 = 2\\sin 3x\ or \y_2 = 2\\cos 3x\?
📖 Explanation: The derivative of \y = A\\sin Bx\ is \AB\\cos Bx\; the derivative of \y = A\\cos Bx\ is \-AB\\sin Bx\. Both attain a maximum magnitude of \|AB|\ when the trigonometric factor equals \\\pm1\. Here \|AB| = |2\\times3| = 6\ for each, so the greatest possible rate of change is the same for both functions.
Q10. Adding \y = A\\sin Bx\ and \y = A\\cos Bx\ yields a resultant amplitude of what?
📖 Explanation: Using the identity \\\sin\\theta + \\cos\\theta = \\sqrt{2}\\sin\\left(\\theta + \\frac{\\pi}{4}\\right)\, the combined function can be written as \\\sqrt{2}\\,A\\sin\\left(Bx + \\frac{\\pi}{4}\\right)\. The coefficient \\\sqrt{2}\\,A\ is the new amplitude, which is larger than the original amplitude \A\ by a factor of \\\sqrt{2}\.
Q11. The point \(A\\sin Bx, A\\cos Bx)\ always lies on which curve?
📖 Explanation: By squaring each coordinate and adding, \(A\\sin Bx)^2 + (A\\cos Bx)^2 = A^2(\\sin^2 Bx + \\cos^2 Bx) = A^2\. This is the equation of a circle centered at the origin with radius \A\. Hence every point generated by the pair lies on that circle.
Q12. In a model of daily temperature using \y = A\\sin Bx\, what do \A\ and \B\ represent?
📖 Explanation: The amplitude \A\ determines how far temperature deviates above or below the average, while \B\ appears inside the sine argument and controls the period via \\\text{period}=\\frac{2\\pi}{B}\. Thus changing \A\ changes the swing of temperature, and changing \B\ changes the length of a day‑cycle.
Q13. If two pendulums start at different times, the phase difference between \y = A\\sin Bx\ and \y = A\\cos Bx\ corresponds to a time offset of what?
📖 Explanation: The cosine wave leads the sine wave by a phase of \\\frac{\\pi}{2}\ radians. Converting phase to time uses the relation \\\text{phase}=B\\times\\Delta t\. Solving \B\\Delta t = \\frac{\\pi}{2}\ gives \\\Delta t = \\frac{\\pi}{2B}\. This offset describes how much later one pendulum begins relative to the other.
Q14. For \y = 5\\sin(\\pi t)\ and \y = 5\\cos(\\pi t)\, at what smallest positive time are the two functions equal?
📖 Explanation: Set \5\\sin(\\pi t) = 5\\cos(\\pi t)\ which simplifies to \\\tan(\\pi t) = 1\. The smallest positive solution to \\\pi t = \\frac{\\pi}{4}\ is \t = \\frac{1}{4}\ seconds, i.e., 0.25 seconds.
Q15. Combining \y = A\\sin Bx\ and \y = A\\cos Bx\ with a coefficient \k\ yields \y = A\\sin(Bx+\\phi)\. What is \\\phi\ in terms of \k\?
📖 Explanation: Writing \A\\sin Bx + kA\\cos Bx = A\\sqrt{1+k^2}\\sin\\left(Bx+\\arctan\\!k\\right)\ uses the identity \\\sin\\alpha+\\tan\\beta\\cos\\alpha = \\sqrt{1+\\tan^2\\beta}\\,\\sin(\\alpha+\\arctan\\!\\tan\\beta)\. Hence the phase shift \\\phi\ required to combine the two terms is \\\arctan(k)\, which determines how the graph is shifted horizontally.
Q16. In the function \y = A\\sin Bx\, what does the parameter \A\ represent?
📖 Explanation: The constant \A\ multiplies the sine term, scaling the output values up or down. This scaling determines the maximum displacement from the horizontal axis, which is precisely the amplitude of the sinusoidal wave.