Question

a) A particle undergoes simple harmonic motion with an angular velocity of 5 rad s-1 and an amplitude of 50 cm. If it starts with maximum forward amplitude at t = 0, find:
i) the displacement at t= 10 s;
ii) the acceleration at t = 6 s.
iii) the velocity at t= 2 s.

b) i) A 5 kg mass undergoes simple harmonic motion on a spring with a force constant of 70 Nm-1. If the amplitude is 3 m, calculate the maximum velocity.

Answer 1

Answer on Question #77240, Physics / Other

a) A particle undergoes simple harmonic motion with an angular velocity of $5\mathrm{rad}\ \mathrm{s}^{-1}$ and an amplitude of $50\mathrm{cm}$. If it starts with maximum forward amplitude at $t = 0$, find:

i) the displacement at $t = 10\ \mathrm{s}$;

ii) the acceleration at $t = 6\ \mathrm{s}$.

iii) the velocity at $t = 2\ \mathrm{s}$.

Solution:

In one-dimensional simple harmonic motion, the position of the particle can be found by

$$x(t) = A \cos(\omega t)$$

Here

$$A = 50\ \mathrm{cm} = 0.5\ \mathrm{m},$$ $$\omega = 5\ \mathrm{rad/s},$$

i) The displacement at $t = 10\ \mathrm{s}$ is

$$x(10) = 0.5 \times \cos(5 \times 10) = 0.482\ \mathrm{m}$$

ii) The acceleration is the 2nd derivative of $x(t)$, or

$$a(t) = -A \omega^2 \cos(\omega t)$$ $$a(6) = -0.5 \times 5^2 \times \cos(5 \times 6) = -10.8\ \mathrm{m/s^2}$$

iii) The velocity is the 1st derivative, or

$$v(t) = -A \omega \sin(\omega t)$$ $$v(2) = -0.5 \times 5 \times \cos(5 \times 2) = -2.46\ \mathrm{m/s}$$

**Answer:** $0.482\ \mathrm{m}; -10.8\ \mathrm{m/s^2}; -2.46\ \mathrm{m/s}$.

b) i) A 5 kg mass undergoes simple harmonic motion on a spring with a force constant of $70\ \mathrm{Nm}^{-1}$. If the amplitude is $3\ \mathrm{m}$, calculate the maximum velocity.

Solution:

The angular velocity is

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{70}{5}} = 3.74\ \frac{\mathrm{rad}}{\mathrm{s}}$$

The maximum values of velocity is

$$v_{\max} = \omega A = 3.74 \times 3 = 11.2\ \mathrm{m/s}$$

**Answer:** $11.2\ \mathrm{m/s}$.

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