Respuesta :
1. 0.42 Hz
The frequency of a simple harmonic motion for a spring is given by:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where
k = 7 N/m is the spring constant
m = 1 kg is the mass attached to the spring
Substituting these numbers into the formula, we find
[tex]f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz[/tex]
2. 2.38 s
The period of the harmonic motion is equal to the reciprocal of the frequency:
[tex]T=\frac{1}{f}[/tex]
where f = 0.42 Hz is the frequency. Substituting into the formula, we find
[tex]T=\frac{1}{0.42 Hz}=2.38 s[/tex]
3. 0.4 m
The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.
4. 0.19 m
We can solve this part of the problem by using the law of conservation of energy. In fact:
- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: [tex]x=0[/tex], so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:
[tex]E=K=\frac{1}{2}mv^2[/tex]
where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass
- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:
[tex]E=U=\frac{1}{2}kA^2[/tex]
Since the total energy must be conserved, we have:
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m[/tex]
5. Amplitude of the motion: 0.44 m
We can use again the law of conservation of energy.
- [tex]E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2[/tex] is the initial mechanical energy of the system, with [tex]x_0=0.4 m[/tex] being the initial displacement of the mass and [tex]v_0=0.5 m/s[/tex] being the initial velocity
[tex]- E_f = \frac{1}{2}kA^2[/tex] is the mechanical energy of the system when x=A (maximum displacement)
Equalizing the two expressions, we can solve to find A, the amplitude:
[tex]\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m[/tex]
6. Maximum velocity: 1.17 m/s
We can use again the law of conservation of energy.
- [tex]E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2[/tex] is the initial mechanical energy of the system, with [tex]x_0=0.4 m[/tex] being the initial displacement of the mass and [tex]v_0=0.5 m/s[/tex] being the initial velocity
[tex]- E_f = \frac{1}{2}mv_{max}^2[/tex] is the mechanical energy of the system when x=0, which is when the system has maximum velocity, [tex]v_{max}[/tex]
Equalizing the two expressions, we can solve to find [tex]v_{max}[/tex], the maximum velocity:
[tex]\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m[/tex]
Answer:
1) The frequency of the simple harmonic motion is [tex]f=0.42\frac{1}{s}[/tex].
The period is [tex]T=2.37s[/tex].
2) If the mass is displaced 0.4 m from its equilibrium position and released from rest, the amplitude of the motion is [tex]A=0.4m[/tex].
3) If the mass is released from its equilibrium position with an initial velocity of 0.5 m/s, the amplitude of the motion is [tex]A=0.19m[/tex].
4) If the mass is displaced 0.4 m from its equilibrium position and released with an initial velocity of 0.5 m/s, the amplitude of the motion is [tex]A=0.44m[/tex] and the maximum velocity is [tex]v_{max} =1.16\frac{m}{s}[/tex].
Explanation:
1) The simple harmonic motion frequency in a system like a mass attached to a spring is determined by the spring constant k, that indicates the stiffness of the spring, and the mass m:
[tex]f=\frac{w}{2\pi} =\frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex]
where [tex]w=\sqrt{\frac{k}{m}}[/tex] is the angular frequency
we are told that [tex]k=7\frac{N}{m}[/tex] and [tex]m=1kg[/tex]
[tex]f=\frac{1}{2\pi}\sqrt{\frac{7N/m}{1kg}}[/tex]
[tex]f=\frac{1}{2\pi}\sqrt{7\frac{1}{s^{2}}}[/tex]
[tex]f=0.42\frac{1}{s}[/tex]
the period T is
[tex]T=\frac{1}{f}[/tex]
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]T=2.37s[/tex]
2) The amplitude is the maximum displacement from equilibrium. If there are no dissipative forces it remains the same throughout the movement. We are told that the mass is displaced 0.4 m from its equilibrium position and released from rest. Then the amplitude of the motion is [tex]A=0.4m[/tex].
3) Depending on the initial conditions we will choose sine or cosine, both periodic, for the expression of displacement as a function of time [tex]x(t)[/tex]. If the mass is displaced a given lenght at t=0 we use cosine. If the mass is at equilibrium position x=0 at t=0 we use sine.
We are told the mass is released from its equilibrium position x=0 at t=0 with an initial velocity of 0.5 m/s. We substitute those values in the expression of velocity that we derive from the expression of displacement as a function of time. We assume the phase to be [tex]\phi=0[/tex].
[tex]x(t)=Asin(wt-\phi)[/tex]
[tex]\frac{dx}{dt}=v(t)=Awcos(wt-\phi)[/tex]
[tex]v(t=0)=Aw=A\sqrt{\frac{k}{m}}[/tex]
[tex]0.5\frac{m}{s} =A\sqrt{7} \frac{1}{s}[/tex]
[tex]A=0.19m[/tex]
4) We are told the mass is displaced 0.4 m from its equilibrium position and released with an initial velocity of 0.5 m/s. The expressions of velocity and of displacement as a function of time are:
[tex]x(t)=Acos(wt-\phi)[/tex]
[tex]v(t)=-Awsin(wt-\phi)[/tex]
if we substitute t=0
[tex]x_{0} =x(t=0)=Acos(\phi)[/tex]
[tex]v_{0}=v(t=0)=-Awsin(\phi)[/tex]
then we use the trigonometric identities [tex]cos(-\phi)=cos(\phi)[/tex] and [tex]sin(-\phi)=-sin(\phi)[/tex]
[tex]x_{0} ^{2}=A^{2}cos(\phi)^{2}[/tex]
[tex]\frac{v_{0} ^{2}}{w^{2}} =A^{2}sin(\phi)^{2}[/tex]
if we add this two expressions we get
[tex]x_{0}^{2}+\frac{v_{0} ^{2}}{w^{2}} =A^{2}(sin(\phi)^{2}+cos(\phi)^{2})[/tex]
we use the trigonometric identity [tex](sin(\phi)^{2}+cos(\phi)^{2})=1[/tex] to get
[tex]A=\sqrt{x_{0}^{2}+\frac{v_{0} ^{2}}{w^{2}}}[/tex][tex]A=\sqrt{(0.4)^{2}m^{2}+\frac{(0.5)^{2}\frac{m^{2}}{s^{2}}}{(\sqrt{7})^{2}\frac{1}{s^{2}}}}[/tex]
[tex]A=\sqrt{(0.4)^{2}+\frac{(0.5)^{2}}{(\sqrt{7})^{2}})m^{2}}[/tex]
[tex]A=0.44m[/tex]
maximum velocity occurs when [tex]sin(wt-\phi)=-1[/tex] in the expression of [tex]v(t)[/tex]
[tex]v_{max}=Aw[/tex]
[tex]v_{max}=0.44m .\sqrt{7} \frac{1}{s}[/tex]
[tex]v_{max} =1.16\frac{m}{s}[/tex]
