6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically. The elastic constant of the string is 500 N/m. The mass is pulled down a vertical distance of 20 cm from the equilibrium position and is then released from rest. (i) Show that the particle moves with simple harmonic motion. (ii) Find the speed and acceleration of the mass 0.1 seconds after it is released from rest. (b) A and B are two fixed pegs, A is 4 m vertically above B. A mass m kg, connected to A and B by two light inextensible strings of equal length, is describing a horizontal circle with uniform angular velocity ω. For what value of ω will the tension in the upper string be double the tension in the lower string?

Leaving Cert Applied Maths Circular Motion: 6. (a) A particle of mass 5 kg i

6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically - Leaving Cert Applied Maths - Question 6 - 2008

Question 6

6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically. The elastic constant of the string is 500 N/m. The m... show full transcript

Worked Solution & Example Answer:6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically - Leaving Cert Applied Maths - Question 6 - 2008

Step 1

Show that the particle moves with simple harmonic motion.

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Answer

Speed and Acceleration Calculation:
- From the previous part, the amplitude of oscillation is given by:
$x = a imes ext{cos}( heta)$

Here, the amplitude a = 0.2 m, so at time t = 0.1 s, we calculate:

$x = 0.2 ext{ cos}( heta)$
- Using $heta = rac{2 ext{π}}{T}t$ , where T is the period of the motion, let’s find velocity:
$v = rac{d}{dt}(a ext{ cos}( heta)) = -a ext{ sin}( heta) rac{d heta}{dt}$

With $d heta/dt = rac{2 ext{π}}{T}$ , we find:

$v = -a rac{2 ext{π}}{T} ext{ sin}( rac{2 ext{π}}{T}t)$

Substituting gives:
- Using values,
- T calculated from acceleration value is:
$T = 2 ext{π} imes ext{sqrt}( rac{m}{k}).$

Substituting values:
- Acceleration is found using:
$a = rac{d^2x}{dt^2}$ gives you acceleration as:

$a = - rac{k}{m}x$

Which gives numerical values leading to:
- Speed at t = 0.1 s is approx. 1.68 m/s
- Acceleration is approximately 10.8 m/s².

Step 2

For what value of ω will the tension in the upper string be double the tension in the lower string?

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Answer

Tension Analysis:

For the horizontal circular motion, we consider the forces acting on the mass:
- Let the tensions in the upper and lower strings be T and T', respectively.
- The radius of the circle traced by the mass is 2 m (since the length of the strings is 4 m total).
The equations from the vertical and horizontal force balances are:
1. From the vertical component:
$2T ext{ cos θ} = mg$
1. From the horizontal component:
$T' ext{ = } rac{mv^2}{r}$

Given the condition where T = 2T', we create a relationship:

Substituting,

$3T = mg$

Then simplifying:

$T = \frac{mg}{3}$

Dual expressing motions develop:

Then arrive at:

$T' = \frac{mg}{6}.$

From radially through motion:
- Applying circular motion, we can use:
$2T sin θ + T' sin θ = mrω^2$

This leads into:

$\frac{3mg}{2} sin θ = m r ω^2$

Solving ultimately resolves to find the specific value of ω through equations leading to:
ω = √\frac{3g}{2}.$$

6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically - Leaving Cert Applied Maths - Question 6 - 2008

Worked Solution & Example Answer:6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically - Leaving Cert Applied Maths - Question 6 - 2008

Show that the particle moves with simple harmonic motion.

For what value of ω will the tension in the upper string be double the tension in the lower string?

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