Use Sinusoidal Functions to Solve Applications
with Solutions
Problems with Solutions

A mass attached to a spring is pulled toward the floor so that its height above the floor is 10 mm (millimeters). The mass is then released and starts moving up and down reaching maximum and minimum heights of 20 and 10 mm , respectively, with a cycle of 0.8 seconds.
a) Assume that the height h(t) of the mass is a sinusoidal function, where t is the time in seconds, sketch a graph of h from t = 0 to t = 0.8 seconds. t = 0 is the time at which the mass is released.
b) Find a sinusoidal function for the height h(t).
c) For how many seconds is the height of the mass above 17 mm over one cycle?
Solution
a) The mass is released at t = 0 when h is minimum. Half a cycle later h reaches its maximum and another half a cycle it reaches its minimum again. Hence over one cycle, h varies with t as follows:
b) According to the graph obtained in part a), h(t) could be modeled by a cosine function shifted (translated) vertically up and horizontally to the right. Hence
h(t) = a cos[ b(t  d) ] + c
Let hmax be the maximum value of h and hmin be the minimum value of h. Hence
a = (hmax  hmin) / 2 = (20  10) / 2 = 5 , a = ~+mn~5
c = (hmax + hmin) / 2 = (20 + 10) / 2 = 15
Period = 2π / b = 0.8, hence b = ~+mn~ 2.5π
We use a = 5 and b = 2.5π. The shift of the cosine function is to the right and equal to half a period. Hence d = 0.4
h(t) = 5 cos[ 2.5π(t  0.4) ] + 15
Check that h has a minimum at t = 0: h(0) = 5 cos[ 2.5π(0  0.4) ] + 15 = 5 cos( π ) + 15 = 10
Check that h has a maximum at t = 0.4: h(0) = 5 cos[ 2.5π(0.4  0.4) ] + 15 = 5 cos( 0 ) + 15 = 20
c) Below is shown the graph of y = h(t) and y = 17. We first need to find t1 and t2 which are the values of t for which h(t) = 17 by solving the equation
5 cos[ 2.5π(t  0.4) ] + 15 = 17
cos[ 2.5π(t  0.4) ] + 15 = (17  15) / 5 = 0.4
2.5π(t  4) = arccos(0.4)
t = arccos(0.4) / (2.5π) + 0.4 = 0.547 seconds
The solution t found above is larger that o.4, which is the position of the maximum, and smaller than 0.8 and it therefore corresponds to t2. Hence
t2 = arccos(0.4) / (2.5π) + 0.4 = 0.547 seconds
t1 is obtained using the symmetry of the two solutions with respect to the position of the maximum to t = 0.4 . Hence
t1 = 0.4  (0.547  0.4) = 0.252 seconds
The height of the mass is more than 17 meters for
t2  t2 = 0.547  0.252 = 0.295 seconds

The number of hours of daylight H in a certain area is approximately given by the function
H(t) = 2.5 cos[ b(t  d) ] + 11.5
where H is in hours and t in days, and the function has a period of one year (365 days).
a) Find b (b > 0) and d if H is maximum on June 21st (month of February has 28 days).
b) Which day is the shortest (has the smallest number of hours of daylight)?
Solution
a) Since the period is known and equal to 365 days, then .
365 = 2π / b , hence b = 2π / 365
If we set d = 0 in the function H(t) = 2.5 cos[ b(t  d) ] + 11.5, it becomes H(t) = 2.5 cos[ b t ] + 11.5 which has a maximum at t = 0.
In our problem, the maximum happens on the 21 st of June corresponding to
t = 31 + 28 + 31 + 30 + 31 + 21 = 172 (number of days from January 1st to 21 st of June)
Therefore the horizontal shift (translation) of 172 is to the right and d = 172. Hence
H(t) = 2.5 cos[ 42.2(t  172) ] + 11.5
b) The shortest day corresponds to t that gives H minimum which is equal to 11.5  2.5 = 9. Hence we need to solve the equation
2.5 cos[ (2π / 365)(t  172) ] + 11.5 = 9
cos[ (2π / 365)(t  172) ] = (9  11.5) / 2.5 =  1
(2π / 365)(t  172) = π
t = 365π/(2π) + 172 = 354.5 days
NOTE We could have answered part b) using the fact that the distance between a maximum and the following minimum in a sinusoidal function is half a period and therefore h is minimum at t = 172 + (1/2)365 = 354.5 days
For the first 11 months (from January to November) of the year, t is given by
t = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 = 334
Number of days in December
354.5  334 = 20.5 number of days in December
which approximately corresponds to the 21 st of December. 
The monthly average temperature T (in °C) in a certain city, may be approximated by
T(t) = a cos[ b(t  d) ] + c
where t is the time in months, t = 0 corresponds to January 1 st, and we assume that function T(t) has a period of 12 months.
a) Find a, b (b > 0), c and d if T has a maximum of 22.4 °C in the middle of July (t = 6.5) and a minimum of  10 °C.
b) In which month is T minimum?
c) Over how many months is T greater than 18 °C?
Solution
a) The minimum and maximum values of T , Tmax and Tmin respectively, allow us to find a and c as follows
We are given: Tmax = 22.4 and Tmin = 10
c = (Tmax + Tmin) / 2 = 6.2
a = (Tmax  Tmin) / 2 = 16.2 , we take a > 0 and equal to 16.2
The graph of T is that of a a cos(bt) shifted 6.5 to the right. Hence
T(t) = 16.2 cos[ b(t  6.5) ] + 6.2
We now use the period to find b.
The period is equal to 12 = 2π / b , hence b = π / 6
T(t) is given by
T(t) = 16.2 cos[ (π / 6)(t  6.5) ] + 6.2
b) Find t for which T is minimum. The minimum value of T is equal to 10. Hence t that makes T minimum is found by solving
16.2 cos[ (π / 6)(t  6.5) ] + 6.2 = 10.2
cos[ (π / 6)(t  6.5) ] =  1
(π / 6)(t  6.5) = π
t = 12.5
NOTE We could have answered part b) using the fact that the distance between a maximum and the following minimum in a sinusoidal function is half a period and therefore T is minimum at t = 6.5 + (1/2)12 = 12.5 days
t = 12.5 is larger than one period which is 12. Hence 12.5 corresponds to a period of 12 months (one year) and 0.5 month of January. So the temperature is minimum in mid January.
c) (See graph below) We first find the times t1 and t2 at which T = 18 by solving the equation
16.2 cos[ (π / 6)(t  6.5) ] + 6.2 = 18
cos[ (π / 6)(t  6.5) ] = (18  6.2) / 16.2
(π / 6)(t  6.5) = arccos((18  6.2) / 16.2)
t = (6/π) arccos((18  6.2) / 16.2) + 6.5 = 7.94 months
Examine the graph of T below and the horizontal line y = 18 (graphical interpretation of the equation above), we note that the solution t = 7.94 is larger that 6.5 which corresponds to t2 = 7.94. By symmetry t1 may be calculated as follows
t1 = 6.5  (t2  6.5) = 6.5  (7.94  6.5) = 5.06 months
The number of months for which T is greater than 18 is equal to
t2  t1 = 7.94  5.06 = 2.88 ; approximately 3 months. 
The diameter of a large Ferris wheel is 48 meters and it takes 2.8 minutes for the wheel to complete one revolution. A rider gets onto the wheel at its lowest point which is 60 cm above ground at t = 0.
a) Find a sinusoidal function h(t) that gives the height h, in meters, of the rider above ground as a function of the time t in minutes.
b) Find the time intervals for which the rider is at a height less than 30 meters for the period of time from t = 0 to t = 2.8 minutes.
c) How many minutes, from t = 0, does it take the rider to reach the highest point for the second time?
Solution
a) The minimum height hmin above the ground is 0.6 meters. The maximum height hmax is equal to the minimum height plus the diameter of the weel.
hmax = 0.6 + 48 = 48.6
Since h(t) is minimum at t = 0, it would be easier to model it by a reflected cos(x) function. Hence
h(t) = a cos [b(t  d)] + c
a = (hmax  hmin) / 2 = (48.6  0.6) / 2 = 24 , two solutions for a = ~+mn~24
We take a = 24 where the minus sign account for the reflection on the horizontal axis.
c = (hmax + hmin) / 2 = (48.6 + 0.6) / 2 = 24.6
The period = 2.8 = 2π/b , hence b = 2π/2.8
h(t) =  24 cos [ (2π/2.8)t ] + 24.6
Check that a t = 0 , h(0) = 0.6 m the minimum height and a t = 1.4 (half a period later) , h(1.4) =  24 cos ( (2π/2.8) 1.4 ) + 24.6 =  24 cos ( π ) + 24.6 = 48.6 m is maximum.
b) We first need to solve the equation
 24 cos [ (2π/2.8)t ] + 24.6 = 30
cos [ (2π/2.8)t ] = (30  24.6) / (24)
t = 2.8 arccos [ (30  24.6) / (24) ] / (2 π) = 0.8 mn
The solution found corresponds to t1 which is the intersection of the graph of h(t) and y = 30.
hence t1 = 0.8 mn and t2 = 2.8  0.8 = 2 mn.
h(t) is less than 30 from t =0 to t = 0.8 mn and from t = 2 to t = 2.8 mn a total of 1.6 mn.
c) The rider reaches the maximum at t = half a period for the first time and t = half a period + one period the second time. Hence it takes
(1/2)2.8 + 2.8 = 4.2 mn for the rider to reach the maximum for the second time. 
Because of the gravitational attractions of the moon and the sun on the Earth, water in seas and oceans tend to rise and fall periodically corresponding to what is called high and low tides. In a a typical situation, the time between two high tides is close to 12 hours. In a certain costal area, the depth of water may be approximated by a sinusoidal function of the form d(t) =  2.5 cos[ b(t  2) ] + 3.5, where d is in meters and t in in hours where t = 0 corresponds to 12 am.
a) Find b (b > 0) if d has a period of 12 hours.
b) From t =0 to t = 12, at what time is d the smallest (low tide) and at what time it is highest (high tide)?
c) From t = 0 to t = 12, what are the interval of time during which the depth of the water 4.5 meters or more?
Solution
a) Using the period, we have
12 = 2π/b
b = π/6
b) d(t) is now given y
d(t) =  2.5 cos[ (π/6)(t  2) ] + 3.5
the smallest value of d = 2.5 + 3.5 = 1
hence d is smallest for t such that  2.5 cos[ (π/6)(t  2) ] + 3.5 = 1
Solve to get : cos[ (π/6)(t  2) ] = 1
(π/6)(t  2) = 0
t = 2 , corresponding to 2 am
the highest value of d = 2.5 + 3.5 = 6
hence d is largest for t such that  2.5 cos[ (π/6)(t  2) ] + 3.5 = 6
Solve to get : cos[ (π/6)(t  2) ] =  1
(π/6)(t  2) = π
t = 8 , corresponding to 10 am
NOTE We could have answered part b) using the fact that the distance between a minimum and the following maximum in a sinusoidal function is half a period and therefore d is maximum at t = 2 + (1/2)12 = 8
c) We first need to find t for which h(t) = 4.5 by solving the equation
 2.5 cos[ (π/6)(t  2) ] + 3.5 = 4.5
t1 = 6 arccos(1/2.5) + 2 = 5.8 hours
t2 = 8 + (8  5.8) = 10.2 hours (use symmetry with respect to position of maximum)
total number of hours for which d(t) is more than 4.5 m is : 10.2  5.8 = 4.4 hours.

Due to the high and low tides, the depth d of water in a certain costal area may be expressed by a sinusoidal function. The highest tide occurs at 8 am and the lowest tide occurs 6 hours later. The maximum level of water is 2.8 meters and the lowest level of water is 0.4 meters.
a) Use sinusoidal functions to find the depth d(t) of the water, in meters, as a function of time t in hours. (Assume that 8 am corresponds to t = 0).
b) Find the depth of water at noon.
c) Use the graph of d(t) and analytical calculations to calculate the interval of time during which the depth d is below 1.5 m from 12 pm to 6 pm.
Solution
a) Let d(t) be written as
d(t) = a cos[b(t  d)] + c
The minimum dmin and the maximum dmax of d are
dmin = 0.4
dmax = 2.8
c = (dmax + dmin) / 2 = (2.8 + 0.4) / 2 = 1.6
a = (dmax  dmin) / 2 = (2.8  0.4) / 2 = 1.2
Since d(t) has a minimum at t = 0 (8 am), we can select a =  1.2 and d = 0
d(t) = 1.2 cos[b(t)] + 1.6 , where it is easy to check that d = 1.2 + 1.6 = 0.4 at t = 0.
We now use the period to find b (b > 0) as follows
period = 12 = 2π / b
hence b = π / 6
d(t) is now written as
d(t) = 1.2 cos[ (π / 6)(t) ] + 1.6 , where it is easy to check that the maximum occurs at t = 6.
b) At noon t = 4, hence
d(4) = 1.2 cos[ (π / 6)(4) ] + 1.6 = 2.2 m
c) The graph of d(t) is shown below with vertical lines corresponding to 12 pm to 6 pm and a horizontal line corresponding to d = 1.5. The depth of the water is less than 1.5 from t_{0} to 6 pm (t = 10). We need to find t_{0} which is a point of intersection of d(t) and y = 1.5 by solving the equation
1.2 cos[ (π / 6)(t) ] + 1.6 = 1.5
t = 6 arccos ( (1.5  1.6)/(1.2) ) / π ≈ 2.84
The solution found corresponds to the left point of intersection of d(t) and y = 1.5. The right point t_{0} may be found using the symmetry of the graph with respect to the maximum point. Hence
t_{0} = 6 + (6  2.84) = 9.16 hours
9.16 hours corresponds to 5 pm and 0.16×60 minutes ≈ 5:10 pm
From 12 pm to 6 pm, the depth of water is below 1.5 from 5:10 pm to 6pm 
A system of solar panels produces a daily average power P that changes during the year. It is maximum on the 21 st of June (day with the highest number of daylight hours) and equal to 20 kwh/day. We assume that P varies with the time t according to the sinusoidal function
P(t) = a cos [b(t  d)] + c
where t = 0 corresponds to the first of January, P is the power in kwh/day and P(t) has a period of 365 days (28 days in February). The minimum value of P is 4 kwh/day.
a) Find the parameters a, b, c and d.
b) Sketch P(t) over one period from t = 0 to t = 365.
c) When is the power produced by the solar system minimum?.
d) The power produced by this solar system is sufficient to power a group of machines if the power produced by the system is greater than than or equal to 16 kwh/day. For how many days, in a year, is the power produced by the system sufficient?
Solution
a) Let P(t) be written as
P(t) = a cos[b(t  d)] + c
The minimum Pmin and the maximum Pmax of P(t) are given.
Pmin = 4
Pmax = 20
c = (Pmax + Pmin) / 2 = (20 + 4) / 2 = 12
a = (Pmax  Pmin) / 2 = (20  4) / 2 = 8
We now need to find the number of days t after January 1st at which P(t) is maximum by counting the days of the months from January to May and adding 21 days in June.
t = 31 + 28 + 31 + 30 + 31 + 21 = 172
We now use the period to find b (b > 0) as follows
period = 365 = 2π / b
hence b = 2π / 365
A cosine function without shift has a maximum at t = 0. P(t) has a maximum at t = 172. We can model P(t) by a cos(x) function shifted by 172 to the right as follows:
P(t) = 8 cos[(2π / 365)(t  172)] + 12
check that P(t) is maximum at t = 172: P(172) = 8 cos[(2π / 365)(172  172)] + 12 = 8 cos[(0)] + 12 = 20
b) The graph of P(t) is shown below.
c) There is a difference of half a period between the maximum at t = 172 and the minimum after that maximum. Hence P(t) is minimum at
t = 172 + 0.5(365) = 354.5
The first 11 eleven months have 334 days. Hence 354.5 corresponds to 21 st of December, day on which P(t) is minimum.
d) To find the number of days for which the power produced is sufficient, we need to find t_{1} and t_{2} corresponding to the points of intersection of P(t) and y = 16 as shown in the graph by solving the equation
P(t) = 16
8 cos[(2π / 365)(t  172)] + 12 = 16
cos[(2π / 365)(t  172)] = 1/2
t = 172 + 365 arccos(1/2) / 2π = 232.8
The solution found above is greater than 172 corresponding to the maximum. Hence the solution above corresponds to t_{2} in the graphs (solution on the right). The solution on the left is found by symmetry as follows:
t_{1} = 172  (232.8  172) = 111.2
t_{2}  t_{1} = 232.8  111.2 = 121.6 days
The system produces enough power for about 121 days.
References
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