Physics Question

Watch this thread
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 1 year ago
#1
A mass suspended from one end of an ideal spring undergoes vertical simple harmonic motion with an amplitude of 2.0cm. Three complete oscillations are made in 4.0s

Find the acceleration of the mass at "equillibrium position"

Find the magnitude of the acceleration of the mass at the position of maximum displacement

Help with this PLZ
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 1 year ago
#2
Acceleration at the equilibrium position is easy - SHM occurs when there is a restoring force pointing towards the equilibrium position AND the magnitude of the restoring force is proportional to the distance from the equilibrium position.
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.

for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units
1
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report Thread starter 1 year ago
#3
(Original post by Joinedup)
Acceleration at the equilibrium position is easy - SHM occurs when there is a restoring force pointing towards the equilibrium position AND the magnitude of the restoring force is proportional to the distance from the equilibrium position.
when the mass is at the equilibrium position the distance from the equilibrium position is zero... and F=ma tells us that the acceleration here must also be zero.

for the other part you will need to select an appropriate formula from the data sheet and convert the information given by the question into the appropriate SI units
Thanks I had another question :

A car of mass 𝑚=1000kg is driven round a smooth circular track of radius 𝑟=250m and takes a time 𝑇=30s to complete one lap. At what angle 𝜃 must the track be banked to counteract the tendency of the car to slip sideways?

?
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 1 year ago
#4
is there a circular motion formula that can help you find the amount of centripetal force for the car?
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report Thread starter 1 year ago
#5
(Original post by Joinedup)
is there a circular motion formula that can help you find the amount of centripetal force for the car?
Centripetal force = mass x velocity2 / radius

But it asks for the angle ?
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report Thread starter 1 year ago
#6
(Original post by Joinedup)
is there a circular motion formula that can help you find the amount of centripetal force for the car?
??helloo
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Report 1 year ago
#7
(Original post by Eesa268)
??helloo
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8
Report Thread starter 1 year ago
#8
(Original post by Joinedup)
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
I’m not sure I understand can you do some working out
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9
Report Thread starter 1 year ago
#9
(Original post by Joinedup)
So what's the angle at which the vertical component of the normal force is equal in size to the weight force and the horizontal component is equal in size to the centripetal force?
can you work it out pls im very confused?
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#10
Report 1 year ago
#10
Have a look at this - you should have done a sketch that looks something like diagram 3 to help you.

https://study.com/academy/lesson/cir...lar-track.html
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#11
Report Thread starter 1 year ago
#11
(Original post by Joinedup)
Have a look at this - you should have done a sketch that looks something like diagram 3 to help you.

https://study.com/academy/lesson/cir...lar-track.html
So am I using the formula Nsintheta=mv^2/R
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#12
Report 1 year ago
#12
(Original post by Eesa268)
So am I using the formula Nsintheta=mv^2/R
That's just a stepping stone... it's got N and sin θ in it - and you don't know N yet but you want θ
you can keep going like it shows till you get
tan θ = v2/(gR)

or you can just look at the sketch diagram you've made which looks something like this
Name:  bankedtrack.jpg
Views: 127
Size:  13.8 KB

and realise that you can get tan θ already because you already know Ny... because it's the size of the weight force on the car
and you already know Nx because it's the centripetal force on the car.

so you can just read off the sketch that tan θ = centripetal force / weight force
and you already know numerical values for both those.

That's why drawing sketches is a really useful thing to remember to do.
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#13
Report Thread starter 1 year ago
#13
f

(Original post by Joinedup)
That's just a stepping stone... it's got N and sin θ in it - and you don't know N yet but you want θ
you can keep going like it shows till you get
tan θ = v2/(gR)

or you can just look at the sketch diagram you've made which looks something like this
Name:  bankedtrack.jpg
Views: 127
Size:  13.8 KB

and realise that you can get tan θ already because you already know Ny... because it's the size of the weight force on the car
and you already know Nx because it's the centripetal force on the car.

so you can just read off the sketch that tan θ = centripetal force / weight force
and you already know numerical values for both those.

That's why drawing sketches is a really useful thing to remember to do.
for v do i use angular speed or normal speed?
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#14
Report 1 year ago
#14
(Original post by Eesa268)
f


for v do i use angular speed or normal speed?
it's the linear speed in m/s, careful about using normal to mean usual cos normal also means perpendicular
in the circular motion types of formulae, angular velocity is always ω (lower case omega)
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#15
Report Thread starter 1 year ago
#15
(Original post by Joinedup)
it's the linear speed in m/s, careful about using normal to mean usual cos normal also means perpendicular
in the circular motion types of formulae, angular velocity is always ω (lower case omega)
I have tried working it out I get very low answer between 0-1
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#16
Report 1 year ago
#16
(Original post by Eesa268)
I have tried working it out I get very low answer between 0-1
OK
circumference of the rack is 2πr = 1570m
speed of the vehicle is d/t = 52.3 m/s

tan θ = v2/(gr)
=2740/(250*9.81)
=1.12

1.12 is the tangent of the angle we're looking for so...
tan-1 1.12 = 48.2°
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#17
Report Thread starter 1 year ago
#17
(Original post by Joinedup)
OK
circumference of the rack is 2πr = 1570m
speed of the vehicle is d/t = 52.3 m/s

tan θ = v2/(gr)
=2740/(250*9.81)
=1.12

1.12 is the tangent of the angle we're looking for so...
tan-1 1.12 = 48.2°
Thanks I was using different formulas to work it out and I didn’t take into account the circumference
0
reply
Eesa268
Badges: 8
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#18
Report Thread starter 1 year ago
#18
(Original post by Eesa268)
Thanks I was using different formulas to work it out and I didn’t take into account the circumference
tan^-1 does not equal 48.2 it is 0.84
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#19
Report 1 year ago
#19
(Original post by Eesa268)
tan^-1 does not equal 48.2 it is 0.84
Can you show what you're doing?
0
reply
Joinedup
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#20
Report 1 year ago
#20
Have you got your calculator set for radians?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest

Were exams easier or harder than you expected?

Easier (32)
27.12%
As I expected (40)
33.9%
Harder (39)
33.05%
Something else (tell us in the thread) (7)
5.93%

Watched Threads

View All