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Physics: Post your doubts here!

Charlotte20102013

Charlotte20102013

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A particle undergoes SHM on a straight line with amplitude 5.0 cm and frequency 2.0 Hz. At time t = 0, it starts to move to the right from the equilibrium position. Determine the instants when the displacements of the object are 1.0 cm and – 2.0 cm for the first time.

I was given the answers 0.02s and 0.28s. But I can't get the answer. Please help.

Thank you! :X3:
 
princess Anu

princess Anu

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shouldn't the current in resistor R be the same as in 50ohm resistor?
 

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Charlotte20102013

Charlotte20102013

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A mass at the end of a helical spring is given a vertical displacement of 3.0cm from its rest position and then released. if the subsequent motion is simple harmonic with a period of 2.0s, find the distance covered by the mass in (i) the first 1.0s (ii) the first 0.75s. [6.0cm; 5.12cm]

Thanks in advance!
 
S

sagar65265

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Charlotte20102013 said:
A mass at the end of a helical spring is given a vertical displacement of 3.0cm from its rest position and then released. if the subsequent motion is simple harmonic with a period of 2.0s, find the distance covered by the mass in (i) the first 1.0s (ii) the first 0.75s. [6.0cm; 5.12cm]

Thanks in advance!
Click to expand...

Okay, so in this problem, the mass in concern is hanging vertically from a spring, and is initially at rest (since it is displaced from it's rest position). It is then displaced either upwards or downwards (doesn't matter which - if the motion is simple harmonic, the amplitude is the same regardless of whether the displacement is upwards or downwards) - for this case, let us assume it is displaced upwards, by pushing it against the spring, a distance of 3.0 cm, after which it starts oscillating.

Here, the amplitude is clearly 3.0 cm - since the amplitude is the distance from the rest position to the point of it's maximum displacement, which in this case is 3.0 cm. Also, if the time period given is 2.0 seconds, the object should, in these 2.0 seconds, travel from the top to the rest position to the bottom, then back to the rest position, and back to the top.

In 1 second, we can assume that the obect completes half an oscillation since each half of the oscillation is identical to the other, and so the mass should divide it's time equally between them. Therefore in half the time of oscillation, the particle travels from the amplitude position to the rest position ( 3.0 cm ) and then from the rest position to the lower amplitude position (3.0 cm), giving us a total of 6.0 cm.

Note that we cannot say this about shorter periods of the oscillation - suppose we take the time when the oscillation begins - it would be absolutely wrong to say that the mass travels as much in the first 10 milliseconds as it does in the 10 milliseconds after that, since the average speed during those time intervals is not the same - however, in this case, we take half the oscillation, and find that the average speed (not velocity - they are asking us to find the distance, not the displacement) is the same in both, so we can divide the particles time equally between them.

For (b), we need to consult the equation of motion for a SHM oscillation:

x(t) = Acos(ωt)

(We use the cosine function since we know that at time t = 0, the mass is at the extreme amplitude position, and at t = 0 in the above cosine function, the value of x is equal to A, the amplitude)

More specifically,

x(t) = (3.0 cm) * cos (ω * t)

We also know that

Angular frequency = ω = 2π/T
where T = Time Period = 2.0 seconds.
So, substituting this into our above equation,

x(t) = (3.0 cm) * cos ([2π/T] * t)

So if we put T = 2.0 seconds and t = 0.75 seconds as per the question's requirements into this equation, we get

x(0.75) = 3.0 cm * cos (3π/4) = -2.12 cm

So what this means is that the mass has gone onto the other side of the rest position and is now 2.12 cm from the rest position - so, the mass has travelled from the upper amplitude point to the rest position (3.0 cm) and has travelled another 2.12 cm onto the other side of the equilibrium position, so we add them to get

3.0 + 2.12 = 5.12 cm.

Here we are adding the modulus of the value of x(0.75) since we want to know the absolute distance it has travelled.
A final note - seeing that the time travelled is 0.75 seconds, we know it has travelled less for than half the period so we should expect an answer less than (a) - however, it is also possible to get this value of -2.12 cm if the particle travelled from the upper amplitude position to the middle (3.0 cm) then to the other amplitude position (3.0 cm) and then upwards until 2.12 cm from the rest point (0.88 cm) thus making the total distance 6.88 cm - so be careful, blindly applying this equation without knowing how the mass moves could get you a wrong answer.

Hope this helped!
Good Luck for all your exams!
 
Charlotte20102013

Charlotte20102013

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sagar65265 said:
Okay, so in this problem, the mass in concern is hanging vertically from a spring, and is initially at rest (since it is displaced from it's rest position). It is then displaced either upwards or downwards (doesn't matter which - if the motion is simple harmonic, the amplitude is the same regardless of whether the displacement is upwards or downwards) - for this case, let us assume it is displaced upwards, by pushing it against the spring, a distance of 3.0 cm, after which it starts oscillating.

Here, the amplitude is clearly 3.0 cm - since the amplitude is the distance from the rest position to the point of it's maximum displacement, which in this case is 3.0 cm. Also, if the time period given is 2.0 seconds, the object should, in these 2.0 seconds, travel from the top to the rest position to the bottom, then back to the rest position, and back to the top.

In 1 second, we can assume that the obect completes half an oscillation since each half of the oscillation is identical to the other, and so the mass should divide it's time equally between them. Therefore in half the time of oscillation, the particle travels from the amplitude position to the rest position ( 3.0 cm ) and then from the rest position to the lower amplitude position (3.0 cm), giving us a total of 6.0 cm.

Note that we cannot say this about shorter periods of the oscillation - suppose we take the time when the oscillation begins - it would be absolutely wrong to say that the mass travels as much in the first 10 milliseconds as it does in the 10 milliseconds after that, since the average speed during those time intervals is not the same - however, in this case, we take half the oscillation, and find that the average speed (not velocity - they are asking us to find the distance, not the displacement) is the same in both, so we can divide the particles time equally between them.

For (b), we need to consult the equation of motion for a SHM oscillation:

x(t) = Acos(ωt)

(We use the cosine function since we know that at time t = 0, the mass is at the extreme amplitude position, and at t = 0 in the above cosine function, the value of x is equal to A, the amplitude)

More specifically,

x(t) = (3.0 cm) * cos (ω * t)

We also know that

Angular frequency = ω = 2π/T
where T = Time Period = 2.0 seconds.
So, substituting this into our above equation,

x(t) = (3.0 cm) * cos ([2π/T] * t)

So if we put T = 2.0 seconds and t = 0.75 seconds as per the question's requirements into this equation, we get

x(0.75) = 3.0 cm * cos (3π/4) = -2.12 cm

So what this means is that the mass has gone onto the other side of the rest position and is now 2.12 cm from the rest position - so, the mass has travelled from the upper amplitude point to the rest position (3.0 cm) and has travelled another 2.12 cm onto the other side of the equilibrium position, so we add them to get

3.0 + 2.12 = 5.12 cm.

Here we are adding the modulus of the value of x(0.75) since we want to know the absolute distance it has travelled.
A final note - seeing that the time travelled is 0.75 seconds, we know it has travelled less for than half the period so we should expect an answer less than (a) - however, it is also possible to get this value of -2.12 cm if the particle travelled from the upper amplitude position to the middle (3.0 cm) then to the other amplitude position (3.0 cm) and then upwards until 2.12 cm from the rest point (0.88 cm) thus making the total distance 6.88 cm - so be careful, blindly applying this equation without knowing how the mass moves could get you a wrong answer.

Hope this helped!
Good Luck for all your exams!
Click to expand...

Thank you so much! (y)
 
guccifier

guccifier

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Hey I need notes on the section B of A2: communication nuclear physics, etc...
I need simple understandable notes
Infact any notes on A2 will be fine.
If any one has access to compiled question and answers please help.
I really want to make an A* for physics.
 
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