How Often Will Two Children Swinging at Different Rates Sync Up in a Minute?

[aknita]

1. Two children are swinging on swings side by side. One completes 20 cycles in 1.0 min while the other completes 26 cycles in 1.0 min. If they start out together, how many additional times will they be swinging together during that minute? Explain.
2. The attempt at a solution
f=N/t
f1=20/60
=.333
f2=26/60
=.433

B.F=f2-f1
=.433-.333
=.1

.1=N/60

N=6 timesthe answer is correct, but i was wondering whether the way i calculated it was the proper method.
thanks in advance.

 

[andrevdh]

Hi aknita,
welcome to PF. Hope we can be of help to you.

The question do not indicate along which lines the problem should be solved. But if it were an exam question the explanation might count more than half of the marks. Personally I would rather use the periods to do the calculation.

So, how did you came to the answer? Can you justify your approach?

 

[m2003]

Yes, your method for calculating the frequency (f) is correct. However, to find the number of additional times they will be swinging together during the minute, you can use the formula N = f x t, where N is the number of cycles, f is the frequency, and t is the time. So, for the first child, N1 = 0.333 x 60 = 20 cycles, and for the second child, N2 = 0.433 x 60 = 26 cycles. Since they start out together, they will be swinging together for 20 cycles in the first minute. However, in the remaining 40 seconds (60 seconds - 20 seconds), the second child will complete an additional 0.433 x 40 = 17.32 cycles. Therefore, the total number of additional times they will be swinging together during that minute is 20 + 17.32 = 37.32, which is approximately 6 times. So, your answer is correct, but the explanation can be improved by using the formula N = f x t.