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Ema1607
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I'm currently doing the June 2013 paper and need help on Q7 (iii) which is simple harmonic motion. I've looked at the mark scheme and still don't get it. I've used the correct equation (x=asin(nt+phase angle)) but don't get why x is -0.2 for the begging part of the solution and why the phase angle is 90 (pi/2) in radians.
The question is:

A particle P of mass m kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 39.2m N. The other end of the string is attached to a fixed point O. The particle is released from rest at O.

(i) Show that, while the string is in tension, the particle performs simple harmonic motion about a point 1 m below O. [3]
(ii) Show that when P is at its lowest point the extension of the string is 0.8 m. [3]
(iii) Find the time after its release that P first reaches its lowest point. [6]
(iv) Find the velocity of P 0.8 s after it is released from O.
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MC11V33N
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(Original post by Ema1607)
I'm currently doing the June 2013 paper and need help on Q7 (iii) which is simple harmonic motion. I've looked at the mark scheme and still don't get it. I've used the correct equation (x=asin(nt+phase angle)) but don't get why x is -0.2 for the begging part of the solution and why the phase angle is 90 (pi/2) in radians.
The question is:

A particle P of mass m kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 39.2m N. The other end of the string is attached to a fixed point O. The particle is released from rest at O.

(i) Show that, while the string is in tension, the particle performs simple harmonic motion about a point 1 m below O. [3]
(ii) Show that when P is at its lowest point the extension of the string is 0.8 m. [3]
(iii) Find the time after its release that P first reaches its lowest point. [6]
(iv) Find the velocity of P 0.8 s after it is released from O.
Hola, I just did that one so I figured I may as well reply:
For part (iii) the reason it has x=-0.2 at the beginning is because, if you remember, the string and mass are in equilibrium at x=1, but l=0.8.
It is important to note that fact that the total time is split into two components - the time for the ball to get from O to 0.8m down (just before there is any tension in the string) and then from 0.8 to the lowest position. The first component is a projectile scenario and suvat will see you through (0.40406 sommet i think).
then you need to get from 0.8 to 1.6, bearing the shm is centred around x=1.
Therefore from an shm perspective, you need to get from -0.2 to 0.6, this is why they have the equation x=Asin(nt) as -0.2=0.6sin(nt).
Theeeeen, if you just rearrange you get (arcsin(-1/3))/7 = t, then you get a negative time so you know you have the wrong part of the sin graph.
If you imagine a sine graph, if you do ^ above you will get 7t=-0.3398, which is in the negative section just left of the y axis, and clearly what you want for a positive t is the corresponding value in the first negative section on the right hand side of the y axis. This translation is pi/2 - (-0.3398) for 7t which gives you 7t= 1.9105, so t=0.273.
Add the two times together, 0.40406+0.273=total time=0.677 s
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Ema1607
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(Original post by MC11V33N)
Hola, I just did that one so I figured I may as well reply:
For part (iii) the reason it has x=-0.2 at the beginning is because, if you remember, the string and mass are in equilibrium at x=1, but l=0.8.
It is important to note that fact that the total time is split into two components - the time for the ball to get from O to 0.8m down (just before there is any tension in the string) and then from 0.8 to the lowest position. The first component is a projectile scenario and suvat will see you through (0.40406 sommet i think).
then you need to get from 0.8 to 1.6, bearing the shm is centred around x=1.
Therefore from an shm perspective, you need to get from -0.2 to 0.6, this is why they have the equation x=Asin(nt) as -0.2=0.6sin(nt).
Theeeeen, if you just rearrange you get (arcsin(-1/3))/7 = t, then you get a negative time so you know you have the wrong part of the sin graph.
If you imagine a sine graph, if you do ^ above you will get 7t=-0.3398, which is in the negative section just left of the y axis, and clearly what you want for a positive t is the corresponding value in the first negative section on the right hand side of the y axis. This translation is pi/2 - (-0.3398) for 7t which gives you 7t= 1.9105, so t=0.273.
Add the two times together, 0.40406+0.273=total time=0.677 s
Thanks so much, the mark scheme really wasn't helping.
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