Question

Suppose a spring-mass system has 18Nm 1 −
k = and m = 0.71kg. The system is
oscillating with an amplitude of 54 mm. (i) Determine the angular frequency of
oscillation. (ii) Obtain an expression for the velocity v of the block as a function of
displacement, x and calculate v at x = 34 mm. (iii) Obtain an expression for the mass’s
distance | x | from the equilibrium position as a function of the velocity v and calculate | x |
when v = 0.18 ms−1. (iv) Calculate the energy of the spring-mass system.

Answer 1

Suppose a spring-mass system has $k = 18Nm^{-1}$ and $m = 0.71kg$. The system is oscillating with an amplitude of $54mm$. (i) Determine the angular frequency of oscillation. (ii) Obtain an expression for the velocity $\upsilon$ of the block as a function of displacement $x$ and calculate $\upsilon$ at $x = 34mm$. (iii) Obtain an expression for the mass's distance $|x|$ from the equilibrium position as a function of the velocity $\upsilon$ and calculate $|x|$ when $\upsilon = 0.18ms^{-1}$. (iv) Calculate the energy of the spring-mass system.

**Solution.**

$$k = 18 \frac{N}{m},\ m = 0.71kg,\ x_{max} = 54mm = 0.054m,\ x = 34mm = 0.034m,\ \upsilon = 0.18\ \frac{m}{s};$$

(i) The angular frequency of oscillation:

$$\omega = \sqrt{\frac{k}{m}}.$$ $$\omega = \sqrt{\frac{18}{0.71}} = 5\left(\frac{rad}{s}\right).$$

(ii) The kinetic energy at the any moment:

$$E_{k} = \frac{mv^{2}}{2}.$$

The potential energy at the any moment:

$$E_{k} = \frac{kx^{2}}{2}.$$

The total amount of energy:

$$E = E_{kmax} = E_{pmax}.$$ $$E = E_{pmax} = \frac{kx_{max}^{2}}{2}.$$

The total amount of energy at the any moment:

$$E = E_{k} + E_{p};$$ $$\frac{kx_{max}^{2}}{2} = \frac{mv^{2}}{2} + \frac{kx^{2}}{2}.$$

An expression for the velocity $\upsilon$ of the block as a function of displacement $x$:

$$\frac{mv^{2}}{2} = \frac{kx_{max}^{2}}{2} - \frac{kx^{2}}{2};$$ $$v ^ {2} = \frac {k}{m} (x _ {m a x} ^ {2} - x ^ {2});$$ $$v = \sqrt {\frac {k}{m} (x _ {m a x} ^ {2} - x ^ {2})}.$$ $$v = \sqrt {\frac {18}{0.71} (0.054 ^ {2} - 0.034 ^ {2})} = 0.21 \left(\frac {m}{s}\right).$$

(iii) The total amount of energy at the any moment:

$$E = E _ {k} + E _ {p};$$ $$\frac {k x _ {m a x} ^ {2}}{2} = \frac {m v ^ {2}}{2} + \frac {k x ^ {2}}{2}.$$

An expression for the mass's distance $|x|$ from the equilibrium position as a function of the velocity $v$ :

$$\frac {k x ^ {2}}{2} = \frac {k x _ {m a x} ^ {2}}{2} - \frac {m v ^ {2}}{2};$$ $$x ^ {2} = x _ {m a x} ^ {2} - \frac {m}{k} v ^ {2};$$ $$x = \sqrt {x _ {m a x} ^ {2} - \frac {m}{k} v ^ {2}}.$$ $$x = \sqrt {0.054 ^ {2} - \frac {0.71}{18} \cdot 0.18 ^ {2}} = 0.04 (m).$$

(iv) The energy of the spring-mass system:

$$E = E _ {p m a x} = \frac {k x _ {m a x} ^ {2}}{2}.$$ $$E = \frac {18 \cdot 0.054 ^ {2}}{2} = 0.026 (J).$$

Answer:

(i) $\omega = 5\frac{rad}{s}.$

(ii) $v(x) = \sqrt{\frac{k}{m}(x_{max}^2 - x^2)}.v(0.034) = 0.21\frac{m}{s}.$

(iii) $x(v) = \sqrt{x_{max}^2 - \frac{m}{k} v^2}. x(0.18) = 0.04m.$

(iv) $E = 0.026J.$