Mixing
Finding the maximum displacement of a bob from its equilibrium position using algebraic/trigonometric methods
The velocity of the bob is
$$v(t)=3sin{t}+4cos{t}$$
At $t=0$, the bob is $1$ unit from the equilibrium position.
I set the derivative equal to zero to find the max displacement, but I got a negative answer ($tapprox-0.927$). $t$ has to be greater than zero, but I don't know how to find the appropriate answer given the one I have.
calculus trigonometry
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
asked Mar 3 '15 at 23:59
user220799
11
add a comment |
The velocity of the bob is
$$v(t)=3sin{t}+4cos{t}$$
At $t=0$, the bob is $1$ unit from the equilibrium position.
I set the derivative equal to zero to find the max displacement, but I got a negative answer ($tapprox-0.927$). $t$ has to be greater than zero, but I don't know how to find the appropriate answer given the one I have.
calculus trigonometry
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
asked Mar 3 '15 at 23:59
user220799
11
2
It asks you to find the maximum DISPLACEMENT, which is the integral of $v(t)$. So you need to find the integral first.
– KittyL
Mar 4 '15 at 0:13
That makes sense, and maybe I explained it wrong but those were the instructions I followed on the problem
– user220799
Mar 4 '15 at 0:31
Is it rectilinear motion or circular? How is equilibrium defined? ( v>0? \omega =0 ). In the latter case what is the string length and position of point of suspensoin to which the bob is connected? Is the angular velocity 1 radian per second?
– Narasimham
May 20 '18 at 17:34
add a comment |
The velocity of the bob is
$$v(t)=3sin{t}+4cos{t}$$
At $t=0$, the bob is $1$ unit from the equilibrium position.
I set the derivative equal to zero to find the max displacement, but I got a negative answer ($tapprox-0.927$). $t$ has to be greater than zero, but I don't know how to find the appropriate answer given the one I have.
calculus trigonometry
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
asked Mar 3 '15 at 23:59
user220799
11
The velocity of the bob is
$$v(t)=3sin{t}+4cos{t}$$
At $t=0$, the bob is $1$ unit from the equilibrium position.
I set the derivative equal to zero to find the max displacement, but I got a negative answer ($tapprox-0.927$). $t$ has to be greater than zero, but I don't know how to find the appropriate answer given the one I have.
calculus trigonometry
calculus trigonometry
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
asked Mar 3 '15 at 23:59
user220799
11
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
asked Mar 3 '15 at 23:59
user220799
11
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
edited Apr 16 '17 at 15:38
Rodrigo de Azevedo
12.8k41855
12.8k41855
asked Mar 3 '15 at 23:59
user220799
11
asked Mar 3 '15 at 23:59
user220799
11
asked Mar 3 '15 at 23:59
user220799
11
11
2
It asks you to find the maximum DISPLACEMENT, which is the integral of $v(t)$. So you need to find the integral first.
– KittyL
Mar 4 '15 at 0:13
That makes sense, and maybe I explained it wrong but those were the instructions I followed on the problem
– user220799
Mar 4 '15 at 0:31
Is it rectilinear motion or circular? How is equilibrium defined? ( v>0? \omega =0 ). In the latter case what is the string length and position of point of suspensoin to which the bob is connected? Is the angular velocity 1 radian per second?
– Narasimham
May 20 '18 at 17:34
add a comment |
2
It asks you to find the maximum DISPLACEMENT, which is the integral of $v(t)$. So you need to find the integral first.
– KittyL
Mar 4 '15 at 0:13
That makes sense, and maybe I explained it wrong but those were the instructions I followed on the problem
– user220799
Mar 4 '15 at 0:31
Is it rectilinear motion or circular? How is equilibrium defined? ( v>0? \omega =0 ). In the latter case what is the string length and position of point of suspensoin to which the bob is connected? Is the angular velocity 1 radian per second?
– Narasimham
May 20 '18 at 17:34
2
2
It asks you to find the maximum DISPLACEMENT, which is the integral of $v(t)$. So you need to find the integral first.
– KittyL
Mar 4 '15 at 0:13
It asks you to find the maximum DISPLACEMENT, which is the integral of $v(t)$. So you need to find the integral first.
– KittyL
Mar 4 '15 at 0:13
That makes sense, and maybe I explained it wrong but those were the instructions I followed on the problem
– user220799
Mar 4 '15 at 0:31
That makes sense, and maybe I explained it wrong but those were the instructions I followed on the problem
– user220799
Mar 4 '15 at 0:31
Is it rectilinear motion or circular? How is equilibrium defined? ( v>0? \omega =0 ). In the latter case what is the string length and position of point of suspensoin to which the bob is connected? Is the angular velocity 1 radian per second?
– Narasimham
May 20 '18 at 17:34
Is it rectilinear motion or circular? How is equilibrium defined? ( v>0? \omega =0 ). In the latter case what is the string length and position of point of suspensoin to which the bob is connected? Is the angular velocity 1 radian per second?
– Narasimham
May 20 '18 at 17:34
add a comment |
2 Answers
2
active
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votes
I am not sure what are the instructions of the problem. I think the displacement can be found by integral, then set the derivative of the displacement to 0 to obtain the maximum. On the other hand, the derivative of the displacement is the velocity itself.
So setting the velocity to 0 would give you the time when the displacement is maximum.
$$3sin{t}+4cos{t}=0 Rightarrow frac{3}{5}sin{t}+frac{4}{5}cos{t}=0\
sin{(t+arccos{(3/5)})}=0$$
So $t=pi-arccos{(3/5)}$. Notice that it is periodic, so using $pi-$ or $2pi-$ are both OK.
Plugging this $t$ into your displacement function would give you the maximum.
An alternative way:
The displacement function is $-3cos{t}+4sin{t}+4$ by integral.
Using the same argument as above this gives you
$$5(-frac{3}{5}cos{t}+frac{4}{5}sin{t})+4$$
The quantity in parenthesis is $sin{(t+theta)}$ for some $theta$. Its maximum must be $1$. So the maximum of the displacement must be $9$.
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
add a comment |
Integrating and noting initial condition we have $x(t)=-3cos(t)+4 sin (t)+4$. This can be written as $x(t)=5cos(t-(pi-arctan(frac{4}{3}))+4$.
Proof:
Consider the vector $vec v=<-3,4>$ and the unit vector $vec w=<cos (theta), sin (theta)>$.
$$vec v cdot vec w=|vec v||vec w| cos( text{angle, between, them})$$
$$=(sqrt{3^2+4^2})(1) cos ((pi-arctan(frac{4}{3}))-theta)$$
$$=(sqrt{3^2+4^2})(1) cos (theta-(pi-arctan(frac{4}{3})))$$
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
add a comment |
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2 Answers
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2 Answers
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votes
I am not sure what are the instructions of the problem. I think the displacement can be found by integral, then set the derivative of the displacement to 0 to obtain the maximum. On the other hand, the derivative of the displacement is the velocity itself.
So setting the velocity to 0 would give you the time when the displacement is maximum.
$$3sin{t}+4cos{t}=0 Rightarrow frac{3}{5}sin{t}+frac{4}{5}cos{t}=0\
sin{(t+arccos{(3/5)})}=0$$
So $t=pi-arccos{(3/5)}$. Notice that it is periodic, so using $pi-$ or $2pi-$ are both OK.
Plugging this $t$ into your displacement function would give you the maximum.
An alternative way:
The displacement function is $-3cos{t}+4sin{t}+4$ by integral.
Using the same argument as above this gives you
$$5(-frac{3}{5}cos{t}+frac{4}{5}sin{t})+4$$
The quantity in parenthesis is $sin{(t+theta)}$ for some $theta$. Its maximum must be $1$. So the maximum of the displacement must be $9$.
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
add a comment |
I am not sure what are the instructions of the problem. I think the displacement can be found by integral, then set the derivative of the displacement to 0 to obtain the maximum. On the other hand, the derivative of the displacement is the velocity itself.
So setting the velocity to 0 would give you the time when the displacement is maximum.
$$3sin{t}+4cos{t}=0 Rightarrow frac{3}{5}sin{t}+frac{4}{5}cos{t}=0\
sin{(t+arccos{(3/5)})}=0$$
So $t=pi-arccos{(3/5)}$. Notice that it is periodic, so using $pi-$ or $2pi-$ are both OK.
Plugging this $t$ into your displacement function would give you the maximum.
An alternative way:
The displacement function is $-3cos{t}+4sin{t}+4$ by integral.
Using the same argument as above this gives you
$$5(-frac{3}{5}cos{t}+frac{4}{5}sin{t})+4$$
The quantity in parenthesis is $sin{(t+theta)}$ for some $theta$. Its maximum must be $1$. So the maximum of the displacement must be $9$.
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
add a comment |
I am not sure what are the instructions of the problem. I think the displacement can be found by integral, then set the derivative of the displacement to 0 to obtain the maximum. On the other hand, the derivative of the displacement is the velocity itself.
So setting the velocity to 0 would give you the time when the displacement is maximum.
$$3sin{t}+4cos{t}=0 Rightarrow frac{3}{5}sin{t}+frac{4}{5}cos{t}=0\
sin{(t+arccos{(3/5)})}=0$$
So $t=pi-arccos{(3/5)}$. Notice that it is periodic, so using $pi-$ or $2pi-$ are both OK.
Plugging this $t$ into your displacement function would give you the maximum.
An alternative way:
The displacement function is $-3cos{t}+4sin{t}+4$ by integral.
Using the same argument as above this gives you
$$5(-frac{3}{5}cos{t}+frac{4}{5}sin{t})+4$$
The quantity in parenthesis is $sin{(t+theta)}$ for some $theta$. Its maximum must be $1$. So the maximum of the displacement must be $9$.
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
I am not sure what are the instructions of the problem. I think the displacement can be found by integral, then set the derivative of the displacement to 0 to obtain the maximum. On the other hand, the derivative of the displacement is the velocity itself.
So setting the velocity to 0 would give you the time when the displacement is maximum.
$$3sin{t}+4cos{t}=0 Rightarrow frac{3}{5}sin{t}+frac{4}{5}cos{t}=0\
sin{(t+arccos{(3/5)})}=0$$
So $t=pi-arccos{(3/5)}$. Notice that it is periodic, so using $pi-$ or $2pi-$ are both OK.
Plugging this $t$ into your displacement function would give you the maximum.
An alternative way:
The displacement function is $-3cos{t}+4sin{t}+4$ by integral.
Using the same argument as above this gives you
$$5(-frac{3}{5}cos{t}+frac{4}{5}sin{t})+4$$
The quantity in parenthesis is $sin{(t+theta)}$ for some $theta$. Its maximum must be $1$. So the maximum of the displacement must be $9$.
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
answered Mar 4 '15 at 10:37
KittyL
13.8k31434
13.8k31434
add a comment |
add a comment |
Integrating and noting initial condition we have $x(t)=-3cos(t)+4 sin (t)+4$. This can be written as $x(t)=5cos(t-(pi-arctan(frac{4}{3}))+4$.
Proof:
Consider the vector $vec v=<-3,4>$ and the unit vector $vec w=<cos (theta), sin (theta)>$.
$$vec v cdot vec w=|vec v||vec w| cos( text{angle, between, them})$$
$$=(sqrt{3^2+4^2})(1) cos ((pi-arctan(frac{4}{3}))-theta)$$
$$=(sqrt{3^2+4^2})(1) cos (theta-(pi-arctan(frac{4}{3})))$$
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
add a comment |
Integrating and noting initial condition we have $x(t)=-3cos(t)+4 sin (t)+4$. This can be written as $x(t)=5cos(t-(pi-arctan(frac{4}{3}))+4$.
Proof:
Consider the vector $vec v=<-3,4>$ and the unit vector $vec w=<cos (theta), sin (theta)>$.
$$vec v cdot vec w=|vec v||vec w| cos( text{angle, between, them})$$
$$=(sqrt{3^2+4^2})(1) cos ((pi-arctan(frac{4}{3}))-theta)$$
$$=(sqrt{3^2+4^2})(1) cos (theta-(pi-arctan(frac{4}{3})))$$
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
add a comment |
Integrating and noting initial condition we have $x(t)=-3cos(t)+4 sin (t)+4$. This can be written as $x(t)=5cos(t-(pi-arctan(frac{4}{3}))+4$.
Proof:
Consider the vector $vec v=<-3,4>$ and the unit vector $vec w=<cos (theta), sin (theta)>$.
$$vec v cdot vec w=|vec v||vec w| cos( text{angle, between, them})$$
$$=(sqrt{3^2+4^2})(1) cos ((pi-arctan(frac{4}{3}))-theta)$$
$$=(sqrt{3^2+4^2})(1) cos (theta-(pi-arctan(frac{4}{3})))$$
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
Integrating and noting initial condition we have $x(t)=-3cos(t)+4 sin (t)+4$. This can be written as $x(t)=5cos(t-(pi-arctan(frac{4}{3}))+4$.
Proof:
Consider the vector $vec v=<-3,4>$ and the unit vector $vec w=<cos (theta), sin (theta)>$.
$$vec v cdot vec w=|vec v||vec w| cos( text{angle, between, them})$$
$$=(sqrt{3^2+4^2})(1) cos ((pi-arctan(frac{4}{3}))-theta)$$
$$=(sqrt{3^2+4^2})(1) cos (theta-(pi-arctan(frac{4}{3})))$$
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
edited Jan 3 '17 at 1:17
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
answered Jan 3 '17 at 0:54
Ahmed S. Attaalla
14.8k12049
14.8k12049
add a comment |
add a comment |
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2
It asks you to find the maximum DISPLACEMENT, which is the integral of $v(t)$. So you need to find the integral first.
– KittyL
Mar 4 '15 at 0:13
That makes sense, and maybe I explained it wrong but those were the instructions I followed on the problem
– user220799
Mar 4 '15 at 0:31
Is it rectilinear motion or circular? How is equilibrium defined? ( v>0? \omega =0 ). In the latter case what is the string length and position of point of suspensoin to which the bob is connected? Is the angular velocity 1 radian per second?
– Narasimham
May 20 '18 at 17:34