The figure below shows two examples of SHM, labeled A and B. (The vertical axis is marked in increments of 0.2 m and the horizontal axis is marked in increments of 0.5 s.)
What are the following quantities for each SHM?
A B
(a) the amplitude 1 m 2 m
(b) the frequency 3 Hz 4 Hz
(c) the period 5 s 6 s
(d) Write the equations for xA and xB in the form of a sine or cosine. Use the following as necessary: t, where x is in meters and t is in seconds.
xA =
xB =
xA = 1sin(2πt/5)
xB = 2sin(2πt/6)
To find the quantities for each SHM labeled A and B, we can use the information provided in the figure.
(a) The amplitude:
The amplitude of an SHM is the maximum displacement from the equilibrium position. In the figure, we can see that the vertical axis is marked in increments of 0.2 m. The maximum displacement above and below the equilibrium position can be measured from the axis to the peak of each wave. For SHM A, the amplitude is 1 m, and for SHM B, the amplitude is 2 m.
(b) The frequency:
The frequency of an SHM is the number of complete oscillations or cycles per unit of time. In the figure, the horizontal axis is marked in increments of 0.5 s. We can count the number of complete oscillations within a given time interval to determine the frequency. For SHM A, there are approximately 3 complete oscillations within 5 seconds, so the frequency is 3 Hz. For SHM B, there are approximately 4 complete oscillations within 6 seconds, so the frequency is 4 Hz.
(c) The period:
The period of an SHM is the time taken for one complete oscillation. It can be calculated as the reciprocal of the frequency. For SHM A, the frequency is 3 Hz, so the period is 1/3 s, which is approximately 0.33 s. For SHM B, the frequency is 4 Hz, so the period is 1/4 s, which is 0.25 s.
(d) The equations for xA and xB in the form of a sine or cosine:
To write the equations for xA and xB, we need to use the general formula for SHM, which is given as:
x = A * sin(2πft + φ)
where x is the displacement from the equilibrium position, A is the amplitude, f is the frequency, t is the time, and φ is the phase constant.
For SHM A:
The amplitude is 1 m, the frequency is 3 Hz, and the phase constant can be determined from the starting position of the wave. Let's assume the wave starts from its peak at t=0. The phase constant, φ, would be 0 degrees or 0 radians. Therefore, the equation for xA is:
xA = 1 * sin(2π * 3t)
For SHM B:
The amplitude is 2 m, the frequency is 4 Hz, and the phase constant can be determined from the starting position of the wave. Let's assume the wave starts from its equilibrium position at t=0. The phase constant, φ, would be 90 degrees or π/2 radians. Therefore, the equation for xB is:
xB = 2 * cos(2π * 4t + π/2)
Note: The choice of sine or cosine function depends on the starting position of the wave. A sine function is used when the wave starts from its equilibrium position, while a cosine function is used when the wave starts from either a peak or a trough.
Therefore, the equations for SHM A and B are as follows:
xA = 1 * sin(2π * 3t)
xB = 2 * cos(2π * 4t + π/2)
(a) The amplitude for SHM A is 1 m, and for SHM B is 2 m.
(b) The frequency for SHM A is 3 Hz, and for SHM B is 4 Hz.
(c) The period for SHM A is 5 s, and for SHM B is 6 s.
(d) To write the equations for xA and xB in the form of a sine or cosine, we can use the general equation for simple harmonic motion:
xA = A * sin(2πft + φ)
xB = A * sin(2πft + φ)
where A is the amplitude, f is the frequency, t is time, and φ is the phase constant.
For SHM A, xA = 1 * sin(2π * 3t + φ)
For SHM B, xB = 2 * sin(2π * 4t + φ)
Note: The phase constant (φ) is not given in the problem, so it cannot be determined without additional information.