Physics: Post your doubts here!

[unique840]

Assalamu alaikum,
This question is about simple harmonic motion:
Q)A small mass attached to a spring oscillates with simple harmonic motion with amplitude of 35mm,taking 6.5s to make 20 oscillations.Calculate
a)it's angular frequencey,
b)it's maximum speed,
c)it's maximum acceleration.
Thanks in advance for your cooperation!

a) 20 oscillations in 6.5 s
1 oscillation in 0.325s
time period = 0.325s
angular frequency = 2pie/time
= 2pie/0.325
= 19.33 rad/seconds
b) maximum speed = ang.frequency*max.amplitude
= 19.33*(35*10^-3)
= 0.67 m/s

c) max acceleration = [(ang.frequency)^2]*max.amplitude
= (19.33^2)*(35*10^-3)
= 13.1 m/s^2

 

[farhan143]

dear brother, i wud like to know where cn i download physics study guide for AS and A level.
Thank you

 

[Mustehssun Iqbal]

a) 20 oscillations in 6.5 s
1 oscillation in 0.325s
time period = 0.325s
angular frequency = 2pie/time
= 2pie/0.325
= 19.33 rad/seconds
b) maximum speed = ang.frequency*max.amplitude
= 19.33*(35*10^-3)
= 0.67 m/s

c) max acceleration = [(ang.frequency)^2]*max.amplitude
= (19.33^2)*(35*10^-3)
= 13.1 m/s^2

Thanks a bunch, really appreciate it.

 

[aliya_zad]

My question concerns (b)(ii). Why don't we consider Rf and R while determining the required value of V2..?

in this case we don't need to consider Rf and R and complicate ourselves.
Vout=Rf/R x( V2-V1)
Vout=0, hence V2-V1=0
V2=V1
So why need to consider Rf and R?
In the third part since Vout is not equal to zero we need to consider Rf and R and in that case they've given us the ratio..

 

[XPFMember]

The link isn't working :'(

As-salam-o-alaikum!

Sorry for the inconvenience....and for the late response..I was a bit busy with my exams..

The links are working now, I've fixed the problem!

 

[XPFMember]

dear brother, i wud like to know where cn i download physics study guide for AS and A level.
Thank you

As-salam-o-alaikum!

The one you're most probably talking about, isn't available on the net!

This one can help you:

 

[unique840]

Thanks a bunch, really appreciate it.

ur welcum

 

[hassam]

in this case we don't need to consider Rf and R and complicate ourselves.
Vout=Rf/R x( V2-V1)
Vout=0, hence V2-V1=0
V2=V1
So why need to consider Rf and R?
In the third part since Vout is not equal to zero we need to consider Rf and R and in that case they've given us the ratio..

ohh kkk i got it....thanks

 

[Mustehssun Iqbal]

Assalamu alaikum

These questions are from CIE Past papers.
Q.21 is from November 1990,Paper 1,Q.5 and Q.33 is from November 1976,Paper 1 and Q.1.
Thanks in advance for your cooperation.

 

[Mustehssun Iqbal]

Sorry,I forgot to add that it's a Force and time graph in Q.21!

 

[unique840]

Assalamu alaikum

These questions are from CIE Past papers.
Q.21 is from November 1990,Paper 1,Q.5 and Q.33 is from November 1976,Paper 1 and Q.1.
Thanks in advance for your cooperation.

q21) momentum = force * time
that means area under graph is momentum
divide the graph in 2 sections
1st division from 0 to 2 seconds, 2nd division from 2 to 6 seconds
area of 1st division : 2*2 = 4
area of 2nd division : 1/2*(4)(2+6) = 16
total area = momentum = 20

 

[Mustehssun Iqbal]

q21) momentum = force * time
that means area under graph is momentum
divide the graph in 2 sections
1st division from 0 to 2 seconds, 2nd division from 2 to 6 seconds
area of 1st division : 2*2 = 4
area of 2nd division : 1/2*(4)(2+6) = 16
total area = momentum = 20

It's very nice of you.Thanks again!

 

[Mustehssun Iqbal]

dear brother, i wud like to know where cn i download physics study guide for AS and A level.
Thank you

I don't know where to get it on the net.Apologies for the late reply.

 

[unique840]

It's very nice of you.Thanks again!

ur welcum again

 

[Mustehssun Iqbal]

Please solve this question from Past papers too:
Q.64) A rod of length 1 metre has non-uniform composition, so that the centre of gravity is not at its geometrical centre.
The rod is laid on supports across two top-pan balances as shown in the diagram.The balances(preiviously set at zero) give readings of 360g and 240g.
Where is the centre of gravity of the rod relative to its geometrical centre? (J93/1/5)
Thanks!

 

[unique840]

option E means what??? whats the numerical value? m getting 0.4. is it ryt?

 

[Mustehssun Iqbal]

option E means what??? whats the numerical value? m getting 0.4. is it ryt?

Apologies for not adding the options too.And it's option A!
A) 1/10 metre to the left
B) 1/10 metre to the right
C) 1/6 metre to the left
D) 1/5 metre to the right
E) 1/5 metre to the left

 

[leosco1995]

I need help with question 9 and 12 of this paper.

Question 12 of this paper.

Question 14 of this paper.

And lastly, question 17 of this paper.


I have a midterm on the 7th, so I'll probably be asking a few more questions along the way. But any help would be appreciated.


http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_12.pdf

 

[abcde]

I need help with question 9 and 12 of this paper.

9. One way of dealing with this question is realising that at the lowest point, the velocity will have a maximum magnitude (min G.P.E and max K.E). Secondly, it will be negative since downward motion is ascribed a negative value. Hence, C is correct.
12. Momentum before collision = momentum after collision
m(-2v) + 3m(v) = 4m V
=> V = mv/4m = v/4.
Hence, A is correct.

 

[unique840]

Apologies for not adding the options too.And it's option A!
A) 1/10 metre to the left
B) 1/10 metre to the right
C) 1/6 metre to the left
D) 1/5 metre to the right
E) 1/5 metre to the left

anticlockwise moments = clockwise moments
assume the distance as 'x'
1 distance will b x metres
the other will b (1-x) metres
360 * x = 240 * (1-x)
360x = 240 - 240x
600x = 240
x = 0.4 metres
as we have to find the center of gravity from the midpoint, it will be 0.5 - x = 0.1 metres to the left