A particle of mass 5 kg is suspended from a fixed point by a light elastic string. The mass is set vibrating vertically with simple harmonic motion, making 4 oscillations per second. (i) Find the constant of elasticity of the string. The mass is now removed from the string. (ii) Find the force required to stretch the string through 1.5 cm from its natural length. (b) A particle of mass m, is suspended by a light inextensible string AB of length 2d. The end A is fixed and the particle is released from rest when AB makes an angle α with the downward vertical through A. When B has risen again to a height so that AB makes an angle of 45° with the downward vertical, the midpoint of the string comes into contact with a small horizontal peg C. (i) If cos α = \( \frac{1}{4} \) find, in terms of d, the speed of the particle at the moment that the string touches the peg. (ii) Find, in terms of m, the tension in the string when the particle reaches the same height as the peg.

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A particle of mass 5 kg is suspended from a fixed point by a light elastic string - Leaving Cert Applied Maths - Question 6 - 2014

Question 6

A particle of mass 5 kg is suspended from a fixed point by a light elastic string. The mass is set vibrating vertically with simple harmonic motion, making 4 oscilla... show full transcript

Worked Solution & Example Answer:A particle of mass 5 kg is suspended from a fixed point by a light elastic string - Leaving Cert Applied Maths - Question 6 - 2014

Step 1

(i) Find the constant of elasticity of the string.

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Answer

To find the elastic constant of the string, we start with the formula for the angular frequency of simple harmonic motion:

\omega = 2\pi f

Where:

( f = 4 ) oscillations per second.

Substituting for ( \omega ):

\omega = 2\pi \times 4 = 8\pi

The formula for angular frequency in terms of the elastic constant ( k ) and mass ( m ) is:

\omega = \sqrt{\frac{k}{m}}

Here, substituting ( m = 5 , \text{kg} ):

8\pi = \sqrt{\frac{k}{5}}

Squaring both sides gives:

(8\pi)^2 = \frac{k}{5} \Rightarrow k = 5(8\pi)^2

Calculating ( k ):

k = 5 \times 64\pi^2 ≈ 3158.27 \, \text{N/m}

Thus, the constant of elasticity is approximately 3158.27 N/m.

Step 2

(ii) Find the force required to stretch the string through 1.5 cm from its natural length.

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Answer

The force required to stretch the string can be calculated using Hooke's Law:

F = kx

Where:

( k ≈ 3158.27 , \text{N/m} )
( x = 1.5 , \text{cm} = 0.015 , \text{m} )

Substituting these values:

F = 3158.27 \times 0.015 ≈ 47.37 \, \text{N}

Thus, the force required to stretch the string through 1.5 cm is approximately 47.37 N.

Step 3

(i) If cos α = \( \frac{1}{4} \) find, in terms of d, the speed of the particle at the moment that the string touches the peg.

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Answer

Using the principle of conservation of energy, the total mechanical energy at the initial height will equal the total mechanical energy when the string touches the peg:

Initial potential energy:

PE_initial = mgh = mg(2d - h)

Where ( h = 2d \cos(α) ). Substitute ( cos(α) = \frac{1}{4} ):

h = 2d(\frac{1}{4}) = \frac{d}{2}

Now substituting for height:

PE_initial = mg(2d - \frac{d}{2}) = mg(\frac{4d - d}{2}) = mg(\frac{3d}{2})

At the point just before touching:

KE + PE = PE_initial

So,

\frac{1}{2}mv^2 + mg \frac{d}{2} = mg \frac{3d}{2}

Solving for ( v ):

v^2 = 2g \frac{3d}{2} - g \frac{d}{2} = 3gd - gd = 2gd

Thus,

v = \sqrt{2gd} ≈ 4.233 \sqrt{d}

Step 4

(ii) Find, in terms of m, the tension in the string when the particle reaches the same height as the peg.

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Answer

At the moment the particle reaches the same height as the peg, applying the forces:

Using the tension formula:

T = mg \cos(45°) + \frac{mv^2}{d}

Since ( v^2 = 2g \frac{d}{2} - g \frac{d}{2} = gd ):

v^2 = gd

Now substituting:

T = mg \cos(45°) + \frac{m(2gd)}{d}

So,

v^2 = 2gd - \sqrt{2}gd

After simplifying:

T = mg \frac{1}{\sqrt{2}} + mg = mg \left(\frac{1}{\sqrt{2}} + 1\right)

This simplifies to yield:

T ≈ 4.06 \, \text{N}

Thus, the tension in the string when the particle reaches the same height as the peg is approximately 4.06 N.

A particle of mass 5 kg is suspended from a fixed point by a light elastic string - Leaving Cert Applied Maths - Question 6 - 2014

Worked Solution & Example Answer:A particle of mass 5 kg is suspended from a fixed point by a light elastic string - Leaving Cert Applied Maths - Question 6 - 2014

(i) Find the constant of elasticity of the string.

(ii) Find the force required to stretch the string through 1.5 cm from its natural length.

(i) If cos α = \( \frac{1}{4} \) find, in terms of d, the speed of the particle at the moment that the string touches the peg.

(ii) Find, in terms of m, the tension in the string when the particle reaches the same height as the peg.

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