1. 0.42 Hz
The frequency of a simple harmonic motion for a spring is given by:
![f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
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
![f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7B7%20N%2Fm%7D%7B1%20kg%7D%7D%3D0.42%20Hz)
2. 2.38 s
The period of the harmonic motion is equal to the reciprocal of the frequency:
![T=\frac{1}{f}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Bf%7D)
where f = 0.42 Hz is the frequency. Substituting into the formula, we find
![T=\frac{1}{0.42 Hz}=2.38 s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7B0.42%20Hz%7D%3D2.38%20s)
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:
, 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:
![E=K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=E%3DK%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
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:
![E=U=\frac{1}{2}kA^2](https://tex.z-dn.net/?f=E%3DU%3D%5Cfrac%7B1%7D%7B2%7DkA%5E2)
Since the total energy must be conserved, we have:
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7DkA%5E2%5C%5CA%3D%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%7Dv%3D%5Csqrt%7B%5Cfrac%7B1%20kg%7D%7B7%20N%2Fm%7D%7D%280.5%20m%2Fs%29%3D0.19%20m)
5. Amplitude of the motion: 0.44 m
We can use again the law of conservation of energy.
-
is the initial mechanical energy of the system, with
being the initial displacement of the mass and
being the initial velocity
is the mechanical energy of the system when x=A (maximum displacement)
Equalizing the two expressions, we can solve to find A, the amplitude:
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx_0%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv_0%5E2%3D%5Cfrac%7B1%7D%7B2%7DkA%5E2%5C%5CA%3D%5Csqrt%7Bx_0%5E2%2B%5Cfrac%7Bm%7D%7Bk%7Dv_0%5E2%7D%3D%5Csqrt%7B%280.4%20m%29%5E2%2B%5Cfrac%7B1%20kg%7D%7B7%20N%2Fm%7D%280.5%20m%2Fs%29%5E2%7D%3D0.44%20m)
6. Maximum velocity: 1.17 m/s
We can use again the law of conservation of energy.
-
is the initial mechanical energy of the system, with
being the initial displacement of the mass and
being the initial velocity
is the mechanical energy of the system when x=0, which is when the system has maximum velocity, ![v_{max}](https://tex.z-dn.net/?f=v_%7Bmax%7D)
Equalizing the two expressions, we can solve to find
, the maximum velocity:
![\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](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx_0%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv_0%5E2%3D%5Cfrac%7B1%7D%7B2%7Dmv_%7Bmax%7D%5E2%5C%5Cv_%7Bmax%7D%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7Dx_0%5E2%2Bv_0%5E2%7D%3D%5Csqrt%7B%5Cfrac%7B7%20N%2Fm%7D%7B1%20kg%7D%280.4%20m%29%5E2%2B%280.5%20m%2Fs%29%5E2%7D%3D1.17%20m%2Fs%20m)