Question

I need a small physics help since the physics section isn't loading for me..please help!



A student says that he had applied force
F=−k√x
on a particle and the particcle moved in SHM..he refuses to tell whether k is a constnt or not..assume that he has worked only wth +ve x a xis and no othe r force acts on the particle..then?
a) x increases as k increases
b) x increases as k decreases
c)x increases ,k=constant
d)indeterminate

Answer 1

attempt:
\[\LARGE y=asinwt\]
\[10=20\sin \frac{2\pi}{T}t\]

Answer 2

144 [s]

Answer 3

2s :P

Answer 4

The mean position is 0cm.

Answer 5

It's at equilibrium at 0cm since it is in the middle of the cycle

Answer 6

Your equation is correct

Answer 7

If you sub in T as 12

Answer 8

then you find t

Answer 9

will it be done like this?
\[\LARGE \frac{1}{2}=tsin \frac{1}{2} \]
t=1s
so total=1*2=2s

Answer 10

but which T as 12s?
there are 2T's right,how are they different

Answer 11

Big T

Answer 12

Big T = 12

Answer 13

what is the small t?

Answer 14

t =1 if we plug it into Wolframalpha

Answer 15

http://www.wolframalpha.com/input/?i=1%2F%282*sin%282*pi%2F12%29%29%3Dt

Answer 16

what does the small t represent?

Answer 17

Lol
Oops

Answer 18

t is inside sine

Answer 19

so what does it mean?

Answer 20

small t represent the time and y represents where the particle is at time t

Answer 21

what does big T represent :S

Answer 22

T represents how long it take for the particle to get back to it original position

Answer 23

So if we start at 0 cm, it will take 12 seconds to get back to 0 cm.

Answer 24

THANKS! :D

Answer 25

Got it?
Here is the answer if you want to check it.
http://www.wolframalpha.com/input/?i=t%3Dasin%281%2F2%29*6%2Fpi

Answer 26

answer is 2s

Answer 27

Yup. It's 2 seconds

Answer 28

yes! twice of that..
can we try another question?

Answer 29

Sure

Answer 30

A particle under the action of SHM has a period of 3second and under the influence of another,it has a time period of 4 seconds..what will be its speed under the combined effect of the two?

Answer 31

I'm not sure what the question means by "under the combined effects of the two".

Answer 32

resultant of the two maybe

Answer 33

The particle slows down since the particle's cycle slows down from 3s to 4s

Answer 34

i have a formula here
\[\LARGE T=\frac{T_{1} T_{2}}{\sqrt{T_{1}^{2}-T_{2}^{2}}} \]

Answer 35

okay leave this one if its unclear !

Answer 36

i don't recognize this equation sorry

Answer 37

Perhaps you can open this question up on a new thread

Answer 38

I dont want to spoil maths section by physics :S

Answer 39

Physic is applied math

Answer 40

lol

Answer 41

:|

Answer 42

2 pendulums have time period T and 5T/4 they start SHM at the same time from mean position..the phse difference between them after the bigger particle has completed 1 oscilation=?

Answer 43

its pendulum and not particle*

Answer 44

do u know this one?

Answer 45

My knowledge in high school physics has limits. :|

Answer 46

okay :o
A student says that he had applied force
\[\LARGE F=-k \sqrt{x}\] on a particle and the particcle moved in SHM..he refuses to tell whether k is a constnt or not..assume that he has worked only wth +ve x a xis and no othe r force acts on the particle..then?

Answer 47

a) x increases as k increases
b) x increases as k decreases
c)x increases ,k=constant
d)indeterminate

Answer 48

\[y=a \sin 2Pift\]
a is amplitude =20cm, f=1/T=1/12 hz, y is the displacement, = 20cm (10cm on each side)
subst into the equation
\[20=20\sin \pi \times 2\times t \div12\]
\[\pi \times t \div 6= \sin^{-1} 1\]
\[\pi t = 1.57 \times 6 = 9.42\]
\[t=9.42\div3.14=3\]
t=3seconds

Answer 49

For the last question, imagine this. You are holding a spring on one hand with your left hand and you are pulling the other end with your right. As you pull further right, you feel a strong force on your right hand to the left. Now let's apply math. Let say x = 0 stands for the spring at equilibrium. As you pull the to the right, x increases positively and remember k is named "the spring constant".

Answer 50

okay

Answer 51

Notice the -k in the equation. There is a negative because the string wants to restore its position whenever it is stretched or compressed.

Answer 52

yes

Answer 53

my answer would be c)x increases ,k=constant

Answer 54

its a :P

Answer 55

k changes?

Answer 56

yes..how

Answer 57

Is this answer from your textbook?

Answer 58

yes

Answer 59

From wikipedia, k is the rate, spring constant or force constant of the spring, a constant that depends on the spring's material and construction

Answer 60

Unless the spring changes its material or its construction, I don't think k will change.

Answer 61

ITS not given that k-constant
read the ques

Answer 62

Hooke's law is \(F=-kx\) not \(-k\sqrt{x}\)

Answer 63

but he assumed thats given in question

Answer 64

we dont have to generalize anything i guess..like we dont even know

Answer 65

if k is a constant or not

Answer 66

check the units

Answer 67

If it's undergoing simple harmonic motion with a single force, that means \(F\propto x\)... i.e. \(-k\propto\sqrt{x}\), so that the force can be written \(F=-k\sqrt{x}=-Kx\) where \(k=K\sqrt{x}\) and therefore non-constant. This is consistent with \(x>0\), so we stay in the reals. It should be clear that since \(k=K\sqrt{x}\), and \(K\) is positive (as the spring constant in \(F=-Kx\)), that an increase in \(x\) translates to an increase in \(k\).

Answer 68

eh?

Answer 69

simple language?

Answer 70

If it's undergoing simple harmonic motion, according to Hooke's law the force must be directly proportional to the displacement i.e. \(F=-Kx\) where \(K\) is our spring constant (always positive). The only force here is \(F=-k\sqrt{x}\), which doesn't immediately resemble our above equation. If we set \(-k\sqrt{x}=-Kx\), you can solve for \(k\) to yield \(k=K\sqrt{x}\). Since \(K\) is positive, and \(x\) is positive, we can see that an increase of \(x\) will cause in an increase in \(\sqrt{x}\) and thus in \(k\).

Answer 71

thanks!