# Homework Help: Vibrations and Pendelums: Work and Answer check

1. Nov 14, 2007

### firefly_1

I have all of the my work and answers, I would just like someone to look over what I've done and make sure I didn't fubar a step along the way. Thanks.

31. Given: A = 1.12 mm; f = 440.0 Hz

Find: $$v_{m}, a_{m}$$

Solution:
$$f = \frac {\omega}{2\pi} \Rightarrow \omega = 2\pi f$$
$$v_{m} = \omega A \Rightarrow 2\pi (440.0Hz)(1.12mm) = 3,096 mm/s = 3.096 m/s$$
$$a_{m} = \omega^{2}A \Rightarrow (2\pi (440.0Hz))^{2}(1.12mm) = 8,560,184 mm/s^{2} = 8560 m/s^{2}$$

32. Given: m = 170 g; f = 3.00 Hz; A = 12.0 cm

Find: a) k (spring constant) b) equation that describes position as a function of time

Solution:
a) $$f = \frac {\omega}{2\pi} = \frac {1}{2\pi} \sqrt {\frac {k}{m}}$$
$$2\pi f^{2} = \sqrt {\frac {k}{m}}^{2} \Rightarrow (m)(2\pi f)^{2} = \frac {k}{m} (m)$$
$$(m)(2\pi f)^{2} = k \Rightarrow (.17kg)(2\pi(3.00Hz))^{2} = 60.4 N/m$$

b) $$y = (12.0cm)sin[(2\pi(3))t] \Leftarrow$$ This is the only one that I am having problems with. This equation is based off of another in the book, but I am not sure if I am even going in the right direction with it.

41. Given: m = 50.0 g; f = 2.0 kHz; A = 1.8 x $$10^{-4}$$m

Find: a) $$F_{m}$$ b) E

Solution:
a) $$a_{m} = \omega^{2}A$$
$$\omega = 2\pi f$$
$$F_{m} = ma_{m}$$
$$= m\omega^{2} A^{2}$$
$$= m (2\pi f)^{2} A^{2}$$
$$= (0.05kg)(2\pi (2 x 10^{3}Hz))^{2}(1.8 x 10^{-4}m)^{2}$$
$$= 1421.22 N = 1.4 kN$$

b) $$v = \omega A$$
$$E = U = \frac {1}{2}mv^{2}$$
$$= \frac {1}{2} m \omega^{2} A^{2}$$
$$= \frac {1}{2}(0.05kg)(2\pi (2 x 10^{3}Hz))^{2}(1.8 x 10^{-4}m)^{2}$$
$$= .13 J [tex] 49. Given: k = 2.5 N/m; y = (4.0cm)sin[(0.70rad/s)t] Find: [tex]K_{m}$$

Solution:
$$v = v_{m} = \omega A$$
$$\omega = \sqrt {\frac {k}{m}}$$
$$K_{m} = \frac {1}{2}mv_{m}^{2}$$
$$= \frac {1}{2}m \omega^{2} A^{2}$$
$$= \frac {1}{2}m (\frac {k}{m}) A^{2}$$
$$= \frac {1}{2} kA^{2} \Rightarrow \frac {1}{2} (2.5N/m)(0.04m)^{2} = .002 J = 2.0 mJ$$

62. Given: L = 120 cm; A = 2.0 cm; E = 5.0 mJ

Find: E if A = 3.0 cm

Solution:
$$v = \omega A$$
$$\omega = \sqrt {\frac {g}{L}}$$
$$E = U = \frac {1}{2} mv^{2}$$
$$= \frac {1}{2} m \omega^{2} A^{2}$$
$$= \frac {1}{2} m (\frac {g}{L}) A^{2}$$
$$= \frac {2EL}{gA^{2}} = m \Rightarrow \frac{2(.005J)(1.2m)}{(9.8m/s^{2})(.02m)^{2}} = 3.06 kg$$
$$E = \frac {1}{2} (3.06kg)(\frac {9.8m/s^{2}}{1.2m})(.03m)^{2} = .0112 J = 11.2 mJ$$

So, that is what I've got. If someone could just let me know how I did, it would be greatly appreciated. Thanks again.