Question

c. What is the acceleration due to gravity on the surface of the planet in m/s2? Determine the number of g-forces.

Answer 1

3. An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph.

Answer 2

We also know that it completes 9 cycles in 5 seconds

Answer 3

What do you think Mohammed?

Answer 4

U hv more than 1 question in 1 post

Answer 5

Anyway i m trying

Answer 6

Do u hv any idea abt g

Answer 7

#3 is not a qustion, it is to let us know the distance

Answer 8

g=gravitational acceleration

Answer 9

Yeah, I assume this is the formula we use
T = 2pi √(length / local gravity)

Answer 10

Correct?

Answer 11

Yes

Answer 12

Both T nd L given

Answer 13

g=?

Answer 14

Well that is what I am confused about. What do we use for gravity? Its not earth but how would we determine gravity in this situation?

Answer 15

We can take square of both sides

Answer 16

Could you write it out?

Answer 17

Because u hv a square root in ur eqiation

Answer 18

T/(2*pi)=√(L/g)

Answer 19

After squaring both sides we will get
(T/(2*pi))^2=L/g

Answer 20

g=?

Answer 21

(5/9(2pi)^2?

Answer 22

What does squaring both sides do for us?

Answer 23

Squaring cancels root

Answer 24

Anyway i m frm cell phone

Answer 25

So unable to write equation nd drawing

Answer 26

Well, whenever you can it would be great. I am going to sleep for now and Ill check back in a while.

Answer 27

After 5 min

Answer 28

Ok Ill wait

Answer 29

now i m frm laptop

Answer 30

\[T=2\pi \times \sqrt{L/g}\]

Answer 31

\[\frac{ T }{ 2\pi }=\sqrt{\frac{ L }{ g }}\]

Answer 32

0.200/2pi = 0.031830

Answer 33

after squaring we will get
\[(\frac{ T }{ 2\pi })^{2}=\frac{ L }{ g }\]

Answer 34

ok

Answer 35

Could you explain what information each equation gives us or what direction it takes us

Answer 36

now
g=L/0.03183

Answer 37

but we need a m/s^2 answer

Answer 38

we can explain the first equation
\[T=2\pi \sqrt{\frac{ L }{ g }}\]

Answer 39

anyway
L is given in ur question

Answer 40

tell me L=?

Answer 41

L= .2 m

Answer 42

correct!!!!!

Answer 43

now put the value of L in ur equation

Answer 44

g=.2/0.03183

Answer 45

g=?

Answer 46

6.2844

Answer 47

6.2833*

Answer 48

correct !!!!!

Answer 49

thank u !!

Answer 50

u can give me a medal if u r satisfied with my explanation

Answer 51

But we need an answer in m/s^2

Answer 52

I also found an answer for this question online but want to understand it better. But here is what I found.

Answer 53

For small swing angles, the period of an ideal pendulum anywhere is

T = 2pi √(length / local gravity) .

The astronaut has already done the pendulum and transmitted the data to us, so
we can use his data and this formula to calculate the local gravity where he is.

P = 2pi √(length / local gravity)

5/9 sec = 2π √(0.2m / gravity)

√(0.2m / gravity) = 5/9sec / 2π

Take the reciprocal of each side: √(gravity) / √(0.2) = 18π / 5

Multiply each side by √(0.2): √(gravity) = 18π √(0.2) / 5

Square both sides: Gravity = (324 π²) (0.2) / 25 = 25.582 m/sec²

Answer 54

unit of acceleration due to gravity is
\[\frac{ m }{ s ^{2} }\]

Answer 55

so the result is different

Answer 56

need to c every steps of our calculation once again

Answer 57

I understand this method until his equation turned into √(gravity) / √(0.2) = 18π / 5

Answer 58

I am not sure how he got 18π

Answer 59

ok

Answer 60

\[\frac{ 5 }{ 9 } \div 2\pi=\frac{ 5 }{ 9 }\times \frac{ 1 }{ 2\pi }=\frac{ 5 }{ 18\pi }\]

Answer 61

response plz for better understanding

Answer 62

Ok, 5/18pi= 0.088419

Answer 63

Where do we go from there?

Answer 64

(T/(2*pi))^2=L/g

If you plug in L=0.2, and T=5/9 into this and then do algebra correctly, you'll have the correct value for g.