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## asib1214 one year ago yelp!!!!!!!!!!!!!!!!!

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1. Michele_Laino

here is my reasoning: |dw:1435602585709:dw| the motion of our object is a periodic motion, and its position goes from x=-9 to x=+9 for example: |dw:1435602699924:dw| so we can write: $\Large 2A = 18$

2. Michele_Laino

A periodic motion, can be described using this function: $\Large x\left( t \right) = A\cos \left( {\omega t + \varphi } \right)$

3. Michele_Laino

now, since: $\Large - 1 \leqslant \cos \left( {\omega t + \varphi } \right) \leqslant 1$ then we have: $\Large - A \leqslant x\left( t \right) \leqslant A$ and by definition, A is the amplitude of our motion

4. Michele_Laino

ok! Then it is suffice that you write that the total length, namely 18 cm, is twice of our ampliude

5. Michele_Laino

amplitude*

6. Michele_Laino

your reasoning is right!

7. Michele_Laino

yor formula is correct!

8. Michele_Laino

your*

9. Michele_Laino

Yes! I think so!

10. Michele_Laino

thanks! :)

11. asib1214

12. asib1214

@sweetburger Hey could you please help me with the question up there in pdg.png

13. asib1214

I've already done the question could you please check my work???!?!?!!?

14. Michele_Laino

the space traveled is 4.8 meters, so the requested velocity is: 4.8/6=0.8 m/sec

15. asib1214

sorry this is how i did it. i've given the distance and frequency =3Hz i can figure out the period. f=1/T f= 1/3 f= 0.33sec/cycle From here i can figure out velocity V= Change in distance/change in time 4.8m/0.33sec 14.5m/sec Now that i have the velocity and frequency, i can certainly figure out the the wavelength by rearranging the formula V=D/t 14.5m/sec all over 3 Hz 4.8m

16. asib1214

@Michele_Laino can you please check my work?!?!?!!? Thanks man

17. asib1214

it's asking me to find the wavelength. Please see the picture i've just uploaded :)

18. Michele_Laino

wavelength is: 4.8/3=... since we have three cycles, within 4.8 meters

19. Michele_Laino

period T= 6/3 = 2 seconds and frequency f= 1/2 = 0.5 Hz

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