[Simple Harmonic Question] I don't how to approach the question (1 Viewer)

[okeydonkey]

Hi could someone please help me with the following question? Thank you so much for the help!
A particle moving in Simple Harmonic Motion starts from rest at a distance 10 metres to the right of its centre of oscillation O. The period of the motion is 2 seconds.
i. Find the speed of the particle when it is 4 metres from its starting point.
i. Find the time taken by the particle to first reach the point 4 metres from its starting point, in seconds correct to 2 decimal places.

 

[hellohi786364]

The general equation for displacement in respect to time for SHM is x = acos(nt+b), a>0, n>0

The particle starts from rest (hence t=0) 10 metres to the right of its centre of oscillation (origin). Note that the particle starts from rest, so this point would be a turning point on its cos graph. Hence, the amplitude is 10 (metres). So a = 10. As the max turning point is at t=0, we also know that b = 0. Period is given as 2, so 2pi/n = 2, giving n a value of pi.

We know know x = 10cos(pi*t)

Velocity is the derivative of displacement, hence x-dot = -10pi*sin(pi*t)

Rearrange our displacement equation to get x/10 = cos(pi*t). cos^2(pi*t) + sin^2(pi*t) = 1, rearrange to get
sin(pi*t) = sqrt(1-cos^2(pi*t)), and substiture cos(pi*t) for x/10 to get sin(pi*t) = sqrt(1-x^2/100)

Now we can substitute in sqrt(1-x^2/100) for sin(pi*t) in our velocity equation to get:
x-dot = -10pi(sqrt(1-x^2/100))


The first question asks for speed when the particle is 4 m from starting point. The particle started 10 metres to the right of the origin at a maximum, therefore we need to find speed when displacement is at 6 m to the right of the origin.

x-dot = -10pi(sqrt(1-6^2/100)) = -8pi

The question only asks for speed, so negative sign is irrelevant. So the answer to the first question is 8pi ms^-1


The second question asks for the time to reach this point, so simply sub in 6 for x in the displacement equation and rearrange to get a result for t as approximately 0.30 seconds.


I have attached an image of the displacement graph (green) and velocity graph (red). The horizontal grey line shows the point where x=6, and the vertical grey line shows the first time that x=6.

 

[okeydonkey]

The general equation for displacement in respect to time for SHM is x = acos(nt+b), a>0, n>0

The particle starts from rest (hence t=0) 10 metres to the right of its centre of oscillation (origin). Note that the particle starts from rest, so this point would be a turning point on its cos graph. Hence, the amplitude is 10 (metres). So a = 10. As the max turning point is at t=0, we also know that b = 0. Period is given as 2, so 2pi/n = 2, giving n a value of pi.

We know know x = 10cos(pi*t)

Velocity is the derivative of displacement, hence x-dot = -10pi*sin(pi*t)

Rearrange our displacement equation to get x/10 = cos(pi*t). cos^2(pi*t) + sin^2(pi*t) = 1, rearrange to get
sin(pi*t) = sqrt(1-cos^2(pi*t)), and substiture cos(pi*t) for x/10 to get sin(pi*t) = sqrt(1-x^2/100)

Now we can substitute in sqrt(1-x^2/100) for sin(pi*t) in our velocity equation to get:
x-dot = -10pi(sqrt(1-x^2/100))


The first question asks for speed when the particle is 4 m from starting point. The particle started 10 metres to the right of the origin at a maximum, therefore we need to find speed when displacement is at 6 m to the right of the origin.

x-dot = -10pi(sqrt(1-6^2/100)) = -8pi

The question only asks for speed, so negative sign is irrelevant. So the answer to the first question is 8pi ms^-1


The second question asks for the time to reach this point, so simply sub in 6 for x in the displacement equation and rearrange to get a result for t as approximately 0.30 seconds.


I have attached an image of the displacement graph (green) and velocity graph (red). The horizontal grey line shows the point where x=6, and the vertical grey line shows the first time that x=6.
View attachment 28389

Hi, thank you so much for your help. How do you know when to use x = acos(nt+b) and x=sin(nt+b) since I was taught both ways. Also,why is that "the particle starts from rest, so this point would be a turning point on its cos graph"? Why must it be the turning point? Thank you!

 

[ultra908]

Hi, thank you so much for your help. How do you know when to use x = acos(nt+b) and x=sin(nt+b) since I was taught both ways. Also,why is that "the particle starts from rest, so this point would be a turning point on its cos graph"? Why must it be the turning point? Thank you!

When the particle starts from rest, that means the velocity at that point is zero, i.e. dx/dt =0, thus a turning point on x(t).
In general, when the particle starts at the centre of motion, use x=sin, when it starts at an extremum, use x=cos