Time Period of a Small Oscillation

[VSayantan]

Homework Statement

[/B]A heavy mass $m$ is suspended from two identical steel wires of length $l$, radius $r$ and Young's modulus $Y$, as shown in the figure above. When the mass is pulled down by a distance $x$ $(x<<l)$ and released, it undergoes elastic oscillations in the vertical direction.
What is the time period?

Homework Equations

Equation of motion of a harmonic oscillator
$$m\frac {d^2 x} {dt^2} = -kx$$
where, $k$ is the spring constant.

Young's modulus of a wire of length $l$, radius $r$ is
$$Y = \frac {FA} {l \cdot \Delta l}$$

Time period
$$T = \frac {2\pi} {\omega}$$

The Attempt at a Solution

[/B]
The vertical components of tension cancel out, while the horizontal components add.
At equilibrium
$$k \cdot l \cdot cos {\frac a 2} = mg$$

With the given small displacement $x<<l$
$$F=k[l \cdot cos {\frac a 2} + x] -mg$$
$$\Rightarrow m \frac {d^2 x} {dt^2} = -kx$$
$$\Rightarrow \frac {d^2 x} {dt^2} = -\omega^2 x$$

Where $\omega = \frac k m = \frac {mg} {m \cdot l \cdot cos {\frac a 2}} =\sqrt {\frac {g} {l \cdot cos {\frac a 2}}}$

So time period
$$T = \frac {2\pi} {\sqrt {\frac {g} {l \cdot cos {\frac a 2}}}} = 2\pi \sqrt {\frac {l \cdot cos {\frac a 2}} {g}}$$

But, I cannot use the expression for the Young's modulus. What am I missing?

 

[Orodruin]

You have assumed the steel wires have zero tension when they have length zero ...

Also, even if that was the case you could not simply assume that the restoring force has the same spring coefficient k.

 

[VSayantan]

You have assumed the steel wires have zero tension when they have length zero ...

Also, even if that was the case you could not simply assume that the restoring force has the same spring coefficient k.

So, if the tension on the wires is $T$ (assuming both of them to be equal, and of course $T\neq0$, as you pointed out), the first equation modifies to $$k \cdot l \cdot cos {\frac a 2}=mg + 2\cdot T \cdot cos {\frac a 2}$$
right?

Then how will I get the equation of motion?

 

[Orodruin]

You have to find the restoring force for a given displacement and use Newton’s second law.

 

[Orodruin]

Also. Not right. You still have some weird kl hanging around your equation. What are you imagining as giving rise to this term and why did you place the tension term on the RHS?

 

[Abhishek kumar]

Homework Statement

View attachment 215364
[/B]A heavy mass $m$ is suspended from two identical steel wires of length $l$, radius $r$ and Young's modulus $Y$, as shown in the figure above. When the mass is pulled down by a distance $x$ $(x<<l)$ and released, it undergoes elastic oscillations in the vertical direction.
What is the time period?

Homework Equations

Equation of motion of a harmonic oscillator
$$m\frac {d^2 x} {dt^2} = -kx$$
where, $k$ is the spring constant.

Young's modulus of a wire of length $l$, radius $r$ is
$$Y = \frac {FA} {l \cdot \Delta l}$$

Time period
$$T = \frac {2\pi} {\omega}$$

The Attempt at a Solution

View attachment 215367[/B]
The vertical components of tension cancel out, while the horizontal components add.
At equilibrium
$$k \cdot l \cdot cos {\frac a 2} = mg$$

With the given small displacement $x<<l$
$$F=k[l \cdot cos {\frac a 2} + x] -mg$$
$$\Rightarrow m \frac {d^2 x} {dt^2} = -kx$$
$$\Rightarrow \frac {d^2 x} {dt^2} = -\omega^2 x$$

Where $\omega = \frac k m = \frac {mg} {m \cdot l \cdot cos {\frac a 2}} =\sqrt {\frac {g} {l \cdot cos {\frac a 2}}}$

So time period
$$T = \frac {2\pi} {\sqrt {\frac {g} {l \cdot cos {\frac a 2}}}} = 2\pi \sqrt {\frac {l \cdot cos {\frac a 2}} {g}}$$

But, I cannot use the expression for the Young's modulus. What am I missing?

So, if the tension on the wires is $T$ (assuming both of them to be equal, and of course $T\neq0$, as you pointed out), the first equation modifies to $$k \cdot l \cdot cos {\frac a 2}=mg + 2\cdot T \cdot cos {\frac a 2}$$
right?

Then how will I get the equation of motion?

Calculate tension in terms of Y,A,L and x.

 

[VSayantan]

Also. Not right. You still have some weird kl hanging around your equation. What are you imagining as giving rise to this term and why did you place the tension term on the RHS?

I'm not sure how to obtain the expression for tension $T$!
Is it, $$T\cdot cos {\frac a 2}=mgx$$?
But it seems that the $cosine$ component of $T$ gets added downward.

Then how will I obtain $k$?

 

[VSayantan]

Calculate tension in terms of Y,A,L and x.

$$Y=\frac {FL} {A\Delta L}$$

Here, $\Delta L=x$, $F=T$
So, $$Y = \frac {FL} {Ax}$$

$$\Rightarrow T= \frac {YAx} {L}$$

Did I get the expression for $F$ correct?

 

[Abhishek kumar]

$$Y=\frac {FL} {A\Delta L}$$

Here, $\Delta L=x$, $F=T$
So, $$Y = \frac {FL} {Ax}$$

$$\Rightarrow T= \frac {YAx} {L}$$

Did I get the expression for $F$ correct?

When system displaced by x then what is elongation in steel rod?

 

[VSayantan]

When system displaced by x then what is elongation in steel rod?

Since the displacement is very small $x<<l$,
$$\frac {x} {\Delta l} \cong cos {\frac a 2}$$

So, $$\Delta l = x\cdot sec {\frac a 2}$$

Is it?

 

[haruspex]

Since the displacement is very small $x<<l$,
$$\frac {x} {\Delta l} \cong cos {\frac a 2}$$

So, $$\Delta l = x\cdot sec {\frac a 2}$$

Is it?

ΔL cannot be more than x. Try again.

 

[VSayantan]

ΔL cannot be more than x. Try again.

Oh!

$$\frac {\Delta l} {x}= cos \frac a 2$$
$$\Rightarrow \Delta l = x \cdot cos \frac a 2$$

Is it okey @haruspex ?

Now using this $\Delta l$ in the expression for Young's modulus, I obtain
$$Y= \frac {Tl} {A \Delta l}$$
$$\Rightarrow T= \frac {Y \pi r^2 x cos \frac a 2} {l}$$

Then I need to equate $T$ to $m\vec a$?

 

[Abhishek kumar]

Oh!

View attachment 215379

$$\frac {\Delta l} {x}= cos \frac a 2$$
$$\Rightarrow \Delta l = x \cdot cos \frac a 2$$

Is it okey @haruspex ?

Now using this $\Delta l$ in the expression for Young's modulus, I obtain
$$Y= \frac {Tl} {A \Delta l}$$
$$\Rightarrow T= \frac {Y \pi r^2 x cos \frac a 2} {l}$$

Then I need to equate $T$ to $m\vec a$?

Yes this is right

 

[Orodruin]

Note that what you have computed is the difference in tension between the equilibrium point and the position $x$, not the actual tension. The tension at the equilibrium point is just enough to cancel the gravitational force and need not be computed. It would be given directly by the geometry of the problem and the weight of the object.

 

[Abhishek kumar]

Note that what you have computed is the difference in tension between the equilibrium point and the position $x$, not the actual tension. The tension at the equilibrium point is just enough to cancel the gravitational force and need not be computed. It would be given directly by the geometry of the problem and the weight of the object.

Yes this is the net restoring force that causes oscillation.

 

[VSayantan]

Note that what you have computed is the difference in tension between the equilibrium point and the position $x$, not the actual tension. The tension at the equilibrium point is just enough to cancel the gravitational force and need not be computed. It would be given directly by the geometry of the problem and the weight of the object.

So, $$m \frac {d^2x} {dt^2}=-\frac {Y \pi r^2 x cos \frac a 2} {l}$$
$$\Rightarrow \frac {d^2x} {dt^2}=-\frac {Y \pi r^2 x cos \frac a 2} {ml}$$

Which gives $$\omega = \sqrt {\frac {Y \pi r^2 cos \frac a 2} {ml}}$$
Then the time period is $$T=\frac {2\pi} {\sqrt {\frac {Y \pi r^2 cos \frac a 2} {ml}}}$$
$$i.e, ~T= 2\pi \cdot \sqrt {\frac {ml} {Y \pi r^2 cos \frac a 2}}$$

Is this right?

 

[Abhishek kumar]

[QUOTE="VSayantan, post: 5886885,

Is this right?[/QUOTE]
Not right.u have taken only force due to one rod into account and motion of mass in vertical direction so you should take component of T in vertical direction.

 

[VSayantan]

Not right.u have taken only force due to one rod into account and motion of mass in vertical direction so you should take component of T in vertical direction.

The total tension on the mass is $2\cdot T \cdot cos \frac a 2$, rather the $x$-component of tension, balanced by the weight $mg$ of the mass.
Then $$mg=2\cdot \frac {Y \pi r^2 x cos {\frac a 2}}{l}\cdot cos \frac a 2$$
Right?
So, how will I get the equation of motion?

 

[Abhishek kumar]

The total tension on the mass is $2\cdot T \cdot cos \frac a 2$, rather the $x$-component of tension, balanced by the weight $mg$ of the mass.
Then $$mg=2\cdot \frac {Y \pi r^2 x cos {\frac a 2}}{l}\cdot cos \frac a 2$$
Right?
So, how will I get the equation of motion?

If you take δl displacement where system is in equilibrium and equating mg with T. Now further displacement x cause oscillation. Write equation for Fnet

 

[VSayantan]

If you take δl displacement where system is in equilibrium and equating mg with T. Now further displacement x cause oscillation. Write equation for Fnet

I'm not sure I can follow your first sentence. $T$ is balanced by $mg$, and not $2T cos \frac a 2$?
So, you mean, $T~+~\Delta T = F_{net}$?
$$i.e., ~mg~+~\frac {2Y\pi r^2 cos^2 {\frac a 2}}{l}=F_{net}$$

Is it?

 

[Abhishek kumar]

I'm not sure I can follow your first sentence. $T$ is balanced by $mg$, and not $2T cos \frac a 2$?
So, you mean, $T~+~\Delta T = F_{net}$?
$$i.e., ~mg~+~\frac {2Y\pi r^2 cos^2 {\frac a 2}}{l}=F_{net}$$

Is it?

2Y*pi*r^2*$l*cos^2(a/2)/l=mg
Now further displacement of x gives
-(2Y*pi*r^2*($l+x)*cos^2(a/2)/l-mg)=ma
Now you can put value of mg and get the result

 

[VSayantan]

2Y*pi*r^2*$l*cos^2(a/2)/l=mg
Now further displacement of x gives
-(2Y*pi*r^2*($l+x)*cos^2(a/2)/l-mg)=ma
Now you can put value of mg and get the result

So, there are actually two equations!
$$2\cdot T \cdot \Delta l \cdot cos \frac a 2= mg$$
and
$$m \frac {d^2x}{dt^2} = -[2\cdot T \cdot (\Delta l + x) \cdot cos \frac a 2 - mg]$$
$$\Rightarrow \frac {d^2x}{dt^2} = - \frac {2\cdot T \cdot x \cdot cos \frac a 2}{m}$$
$$\Rightarrow \frac {d^2x}{dt^2} = - \frac {2Y \pi r^2 cos^2 \frac a 2}{ml}x$$

Hence, $$\omega =\sqrt {\frac {2Y \pi r^2 cos^2 \frac a 2}{ml}}$$
So, time period $$T={\frac {2\pi}{r}} \sqrt {\frac {ml} {2Y \pi cos^2 \frac a 2}}$$

But the thing is there should be no $\pi$ in the denominator of the expression for time period.

 

[Abhishek kumar]

So, there are actually two equations!
$$2\cdot T \cdot \Delta l \cdot cos \frac a 2= mg$$
and
$$m \frac {d^2x}{dt^2} = -[2\cdot T \cdot (\Delta l + x) \cdot cos \frac a 2 - mg]$$
$$\Rightarrow \frac {d^2x}{dt^2} = - \frac {2\cdot T \cdot x \cdot cos \frac a 2}{m}$$
$$\Rightarrow \frac {d^2x}{dt^2} = - \frac {2Y \pi r^2 cos^2 \frac a 2}{ml}x$$

Hence, $$\omega =\sqrt {\frac {2Y \pi r^2 cos^2 \frac a 2}{ml}}$$
So, time period $$T={\frac {2\pi}{r}} \sqrt {\frac {ml} {2Y \pi cos^2 \frac a 2}}$$

But the thing is there should be no $\pi$ in the denominator of the expression for time period.

Why not is there any reason behind that

 

[VSayantan]

Why not is there any reason behind that

Not really.
I've opted for it as the closest answer, though.

Anyway, thanks for your time @Abhishek kumar .

 

[haruspex]

Not really.
I've opted for it as the closest answer, though.

Are you saying that the official answer is the same as yours but without the √π in the denominator?
That must be there since it comes from the cross-sectional area of the wire.

 

[VSayantan]

Are you saying that the official answer is the same as yours but without the √π in the denominator?
That must be there since it comes from the cross-sectional area of the wire.

There are four choices given -
$1. {\frac {2\pi}{r}} \sqrt {\frac {ml} {2\cdot Y \cdot cos^2 {\frac a 2}}}$

$2. 2\pi {\sqrt {\frac {l \cdot cos {\frac a 2}}{g}}}$

$3. \sqrt {\frac {2\pi m l}{Yr^2}}$

$4. {\frac {2\pi} {r}}{\sqrt {\frac {m \cdot g \cdot l} {2Y}}}$

 

[haruspex]

There are four choices given -
$1. {\frac {2\pi}{r}} \sqrt {\frac {ml} {2\cdot Y \cdot cos^2 {\frac a 2}}}$

$2. 2\pi {\sqrt {\frac {l \cdot cos {\frac a 2}}{g}}}$

$3. \sqrt {\frac {2\pi m l}{Yr^2}}$

$4. {\frac {2\pi} {r}}{\sqrt {\frac {m \cdot g \cdot l} {2Y}}}$

You can eliminate all options except (1) by considering how they depend on the variables. g cannot be relevant, the angle a must be, with the period tending to infinity as the angle tends to π, etc.
And as I posted, there must be a π in the frequency because of the dependence on the cross sectional area of the wire, so your answer must be correct.