Shm question -- a mass hanging on a spring vertically set into motion

[jiboom]

A mass on the end of a spring which is hanging vertically is raised up and let go. It then oscillates between 2m and 1.5m above the floor and completes 32 cycles in one minute. The height, h metres, of the mass above the floor after t seconds can be modeled by the function h=acos(pi t / 180) +c where a, b and c are constants.

(a) Determine exactly the period, T, of the oscillation in seconds per cycle and hence find the value of b.

(b) By considering the extremes of the oscillation, work out the values of a and c.

(c) Calculate exactly the value of h when t = 25 seconds.

(d) Find the first time when h = 1.75 metres.

(e) Sketch the graph of h against t for , labelling axes and critical values carefully.
Can I just check my answers please
For first part I have said frequency is 32/60 per sec
Period is 2pi/pi b/180 to give me b
Then I have said this is just shm moved up distance c so
Center is 2+1.5 /2 =1.75
Amplitude is .25 so A is .25
And at t=0 H=2=A+C giving me C

 

[haruspex]

h=acos(pi t / 180) +c where a, b and c are constants.

You missed b. Is it h=acos(pi b t / 180) +c?

For first part I have said frequency is 32/60 per sec
Period is 2pi/pi b/180 to give me b

OK, but giving you what value for b?

Center is 2+1.5 /2 =1.75
Amplitude is .25 so A is .25

Yes.

 

[jiboom]

You missed b. Is it h=acos(pi b t / 180) +c?

OK, but giving you what value for b?

Yes.

Yes the b should be as in your reply.

To find b ill use frequency is 1/period
Are my conditions correct, ie H is 2 at t is 0?
Thanks

 

[haruspex]

Yes the b should be as in your reply.

To find b ill use frequency is 1/period
Are my conditions correct, ie H is 2 at t is 0?
Thanks

Yes.

 

[HalfLostAndSearching]

=1.75

(a) The period, T, of the oscillation can be calculated by dividing the total time of one minute (60 seconds) by the number of cycles completed in that time (32 cycles). This gives us a period of 1.875 seconds per cycle. To find the value of b, we can use the formula T=2pi/(pi*b/180) and solve for b. This gives us b=120 degrees.

(b) The maximum height, or amplitude, of the oscillation can be found by taking the average of the highest and lowest points, which is (2+1.5)/2 = 1.75 meters. This gives us the value of a. To find the value of c, we can use the formula h=acos(pi t / 180) +c and plug in the values for a and the highest point for h (2 meters). This gives us c=0.25 meters.

(c) To calculate the value of h when t=25 seconds, we can plug in the value of t into the equation h=acos(pi t / 180) +c. This gives us h=acos(25pi/180)+0.25=1.75 meters.

(d) To find the first time when h=1.75 meters, we can set h equal to 1.75 in the equation h=acos(pi t / 180) +c and solve for t. This gives us t=25 seconds.

(e) The graph of h against t would be a sinusoidal curve with a center at (0, 1.75) and an amplitude of 0.25. The critical values would be at t=0, t=1.875 seconds, and t=3.75 seconds. The x-axis would represent time in seconds and the y-axis would represent height in meters.