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Period of sine function

A full cycle of “2π” is used to build a sine curve (lower figure).

Elements of periodic motion

Here, we describe certain important attributes of periodic motion, which are extensively used to describe a periodic motion. Though, it is expected that readers are already familiar with these terms, but we present the same for the sake of completeness.

1: Time period (T) : It is the time after which a periodic motion (i.e. the pattern of motion) repeats itself. Its dimensional formula is [T] and unit is “second” in SI unit.

2: Frequency (n, ν) : It is the number of times the unit of periodic motion is repeated in unit time. In SI system, its unit is s - 1 , which is known as "Hertz" or "Hz" in short. An equivalent name is cycles per second (cps). Its dimensional formula is [ T - 1 ]. Time period and frequency are inverse to each other :

T = 1 ν

3: Angular frequency (ω) : It is the product of “2π” and frequency “ν”. In the case of rotational motion, angular frequency is equal to the angle (radian) described per unit time (second) and is equal to the magnitude of average angular velocity.

ω = 2 π ν

The unit of angular frequency is radian/s. In general, angular frequency and angular velocity are referred in equivalent manner. However, we should emphasize that we refer only the magnitude of angular velocity, when quoted to mean frequency. Also,

ω = 2 π T

Since “2π” is a constant, the dimensional formula of angular frequency is same as that of frequency i.e. [ T - 1 ].

4: Displacement (x or y) : It is equal to change in the physical quantity in periodic motion. This physical quantity can be any thing like displacement, electric current, pressure etc. The unit of displacement obviously depends on the physical quantity under consideration.

Period of periodic motion

There are many different periodic functions. In our course, however, we shall be dealing mostly with trigonometric functions. Some important results about period are useful in finding period of a given function.

1: All trigonometric functions are periodic.

2: The periods of sine, cosine, secant and cosecant functions are “2π”, whereas periods of tangent and cotangent functions are “π”.

3: If "k","a" and "b" are positive real values and “T” be the period of periodic function “f(x)”, then :

  • "kf(x)" is periodic with period “T”.
  • "f(x+b)" is periodic with period “T”.
  • "f(x) + a" is periodic with period “T”.
  • "f(ax±b)" is periodic with a period “T/|a|”.

4: If “a” and “b” are non-zero real number and functions g(x) and h(x) are periodic functions having periods, “ T 1 ” and “ T 2 ” , then function f x = a g x ± b h x is also a periodic function. The period of f(x) is LCM of “ T 1 ” and “ T 2 ”.

The fourth LCM rule is subject to certain restrictions. For complete detail read module titled “Periodic functions”.

Examples

Problem 2: Find the time period of the motion, whose displacement is given by :

x = cos ω t

Solution : We know that period of cosine function is “2π”. We also know that if “T” is the period of “f(t)”, then period of the function "f(at±b)" is “T/|a|”. Following this rule, time period of the given cosine function is :

T = 2 π ω

Note that this expression is same that forms the basis of definition of angular frequency.

Problem 3: Find the time period of the motion, whose displacement is given by :

x = 2 cos 3 π t + 5

Solution : We know that period of cosine function is “2π”. If “T” is the time period of f(t), then period of kf(t) is also “T”. Thus, coefficient “2” of trigonometric term has no effect on the period. We also know that if “T” is the period of “f(t)”, then period of the function "f(at±b)" is “T/|a|”. Following this rule, time period of the given cosine function is :

T = 2 π 3 π = 2 3

Problem 4: Find the time period of the motion, whose displacement is given by :

x = sin ω t + sin 2 ω t + sin 3 ω t

Solution : We know that period of sine function is “2π”. We also know that if “T” is the period of “f(t)”, then period of the function "f(at±b)" is “T/|a|”. Following this rule, time periods of individual sine functions are :

Time period of "sin ωt" , T 1 = T ω = 2 π ω

Time period of "sin 2ωt" , T 2 = T 2 ω = 2 π 2 ω = π ω

Time period of "sin 3ωt" , T 3 = T 3 ω = 2 π 3 ω = 2 π 3 ω

Applying LCM rule, we can find the period of combination. Now, LCM of fraction is obtained as :

T = LCM of numerators HCF of denominators = LCM of “2π”, “π” and “2π” HCF of “ω”, “ω” and “3ω”

T = 2 π ω

It is intuitive to understand that the frequencies of three functions are in the proportion 1:2:3. Their time periods are in inverse proportions 3:2:1. It means that by the time first function completes a cycle, second function completes two cycles and third function completes three cycles. This means that the period of first function encompasses the periods of remaining two functions. As such, time period of composite function is equal to time period of first function.

Period of function

The period of first function encompasses the periods of remaining two functions.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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David Reply
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Oscillation and wave motion. OpenStax CNX. Apr 19, 2008 Download for free at http://cnx.org/content/col10493/1.12
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