Question

A 0.300 kg oscillator has a speed of 98.4 cm/s when its displacement is 4.00 cm and 81.4 cm/s when its displacement is 6.00 cm. What is the oscillator's maximum speed?

Answer 1

Answer:

$v_{max}=1.101\ m.s^{-1}$

Explanation:

Given:

• mass of the oscillator, $m=0.3\ kg$

• first case of displacement, $x_1=0.04\ m$

• velocity in the first case, $v_1=0.984\ m.s^{-1}$

• second case of displacement, $x_2=0.06\ m$

• velocity in the second case, $v_2=0.814\ m.s^{-1}$

Now as we know that the total energy in both the cases will remain conserved:

$TE_1=T_2$

$KE_1+PE_1=KE_2+PE_2$

$\frac{1}{2} \times m.v_1^2+\frac{1}{2} \times k.x_1^2=\frac{1}{2} \times m.v_2^2+\frac{1}{2} \times k.x_2^2$

$\frac{1}{2} \times 0.3\times 0.984^2+\frac{1}{2} \times k\times 0.04^2=\frac{1}{2} \times 0.3\times 0.814^2+\frac{1}{2} \times k\times 0.06^2$

$0.1452+k\times 0.0008=0.0994+k\times 0.0018$

$k=45.8\ N.m^{-1}$

Now the total energy:

$TE_1=\frac{1}{2} \times 0.3\times 0.984^2+\frac{1}{2} \times k\times 0.04^2$

$TE_1=\frac{1}{2} \times 0.3\times 0.984^2+\frac{1}{2} \times 45.8\times 0.04^2$

$TE_1=0.1819\ J$

When the whole of the spring potential converts into kinetic energy:

$TE_1=\frac{1}{2} m.v_{max}^2$

$0.1819=\frac{1}{2} \times 0.3\times v_{max}^2$

$v_{max}=1.101\ m.s^{-1}$

Answer 2

The maximum speed of the oscillator is 1.1 m/s.

The given parameters:

• mass of the oscillator, m = 0.3 kg
• speed of the oscillator, v = 98.4 cm/s = 0.984 m/s
• displacement, x₁ = 4.0 cm = 0.04 m
• second displacement, x₂ = 0.06 m
• second speed, v₂ = 81.4 cm/s = 0.814 m/s

Apply the principle of conservation of energy to determine the spring constant;

$K.E_1 + U_1 = K.E_2 + U_2\\\\\frac{1}{2}mv_1^2 + \frac{1}{2} kx_1^2 = \frac{1}{2} mv_2^2 + \frac{1}{2} kx_2^2\\\\mv_1^2 + kx_1^2 = mv_2^2 + kx_2^2\\\\mv_1^2 - mv_2^2 = kx_2^2 - kx_1^2\\\\m(v_1^2 -v_2^2) = k(x_2^2 - x_1^2)\\\\k = \frac{m(v_1^2 -v_2^2)}{x_2^2 - x_1^2} \\\\k = \frac{0.3(0.984^2 \ - \ 0.814^2)}{0.06^2\ - \ 0.04^2} \\\\k = 45.85 \ N/m$

The total energy is calculated as follows;

$M.A = K.E_1 + U_1\\\\M.A = \frac{1}{2} mv_1^2 + \frac{1}{2} kx_1^2\\\\M.A = (0.5)(0.3)(0.984)^2 \ + \ (0.5)(45.85)(0.04)^2\\\\M.A = 0.182 \ J$

The maximum speed of the oscillator is calculated as follows;

$K.E = \frac{1}{2} mv_{max}^2 \\\\v_{max} = \sqrt{\frac{2K.E}{m} } \\\\v_{max} = \sqrt{\frac{2\times 0.182}{0.3} }\\\\v_{max} = 1.1 \ m/s$

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