Answer to Question #260740 in Physics for inchet

Question

A mass m = 4.5 kg is attached to a vertical spring with k = 200 N/m and is set into motion. (a) What is the frequency of the oscillation? (b) If the amplitude of the oscillation is 3.5 cm, what is the maximum value of the velocity? (c) How long does it take the mass to move from y = 1.5 cm to y = 2.5 cm? (d) If the mass is oscillating with a maximum speed of 45 m/s, what is the amplitude? (e) If the spring constant is increased by a factor of two and the maximum kinetic energy of the mass is the same, by what factor does the amplitude change?

Answer 1

(a) The frequency of the oscillation can be found as follows:

$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}},$$ $$f=\dfrac{1}{2\pi}\sqrt{\dfrac{200\ \dfrac{N}{m}}{4.5\ kg}}=1.06\ Hz.$$

(b) The maximum value of the velocity can be found as follows:

$$v_{max}=A\omega=A2\pi f=0.035\ m\times2\pi\times1.06\ Hz=0.23\ \dfrac{m}{s}.$$

(c) Let's write $y_1$ and $y_2$:

$$y_1=Asin\omega t_1,$$ $$y_2=Asin\omega t_2.$$

From these equations we can find $t_1$ and $t_2$:

$$t_1=\dfrac{sin^{-1}(\dfrac{y_1}{A})}{\omega}=\dfrac{sin^{-1}(\dfrac{y_1}{A})}{2\pi f},$$ $$t_2=\dfrac{sin^{-1}(\dfrac{y_2}{A})}{\omega}=\dfrac{sin^{-1}(\dfrac{y_2}{A})}{2\pi f}.$$

Finally, we can find the time that the mass takes to move from y = 1.5 cm to y = 2.5 cm:

$$t=t_2-t_1=\dfrac{sin^{-1}(\dfrac{y_2}{A})-sin^{-1}(\dfrac{y_1}{A})}{2\pi f},$$ $$t=\dfrac{sin^{-1}(\dfrac{0.025\ m}{0.035\ m})-sin^{-1}(\dfrac{0.015\ m}{0.035\ m})}{2\pi\times1.06\ Hz}=0.053\ s.$$

(d) We can find the amplutude as follows:

$$v_{max}=A\omega,$$ $$A=\dfrac{v_{max}}{\omega}=\dfrac{v_{max}}{2\pi f}=\dfrac{45\ \dfrac{m}{s}}{2\pi\times1.06\ Hz}=6.75\ m.$$

(e) We know that the maximum kinetic energy of the mass is the same. It means that $v_{max}=const$. Then, we can write:

$$v_{max}=A\sqrt{\dfrac{k}{m}},$$ $$A=\dfrac{v_{max}}{\sqrt{\dfrac{k}{m}}}.$$

If the spring constant is increased by a factor of two, we get:

$$A_{new}=\dfrac{v_{max}}{\sqrt{\dfrac{2k}{m}}}=A=\dfrac{v_{max}}{\sqrt{2}\sqrt{\dfrac{k}{m}}}=\dfrac{1}{\sqrt{2}}A.$$

Therefore, the amplitude changes by the factor of $\dfrac{1}{\sqrt{2}}.$