Question

Find the resultant amplitude of the following simple harmonic equations:

$x_1 = 5\sin \omega t$

$x_2 = 5\sin(\omega t + 53^\circ)$

$x_3 = -10\cos \omega t$

Answer 1

10

Answer 2

To find the resultant amplitude of the given simple harmonic equations, we can use the method of phasor addition. Each simple harmonic motion (SHM) can be represented by a phasor, which is a vector rotating in a 2D plane with angular velocity $\omega$. The length of the phasor represents the amplitude of the SHM, and the angle it makes with a reference axis at $t=0$ represents its initial phase.

The given equations are: $x_1 = 5\sin \omega t$ $x_2 = 5\sin(\omega t + 53^\circ)$ $x_3 = -10\cos \omega t$

We can express each equation in the standard form $A\sin(\omega t + \phi)$.

1. $x_1 = 5\sin \omega t$: This is already in the standard form with amplitude $A_1 = 5$ and initial phase $\phi_1 = 0^\circ$.

2. $x_2 = 5\sin(\omega t + 53^\circ)$: This is also in the standard form with amplitude $A_2 = 5$ and initial phase $\phi_2 = 53^\circ$.

3. $x_3 = -10\cos \omega t$: We need to convert $-\cos \omega t$ into a sine function. We know that $-\cos \theta = \sin(\theta - 90^\circ)$ or $-\cos \theta = \sin(\theta + 270^\circ)$. Using the phase $-90^\circ$: $x_3 = 10(-\cos \omega t) = 10\sin(\omega t - 90^\circ)$. So, the amplitude is $A_3 = 10$ and the initial phase is $\phi_3 = -90^\circ$.

Now we have three phasors with the following amplitudes and initial phases: Phasor 1: $A_1 = 5$, $\phi_1 = 0^\circ$ Phasor 2: $A_2 = 5$, $\phi_2 = 53^\circ$ Phasor 3: $A_3 = 10$, $\phi_3 = -90^\circ$

The resultant displacement is the sum of the individual displacements: $x = x_1 + x_2 + x_3$. The resultant motion is also an SHM with the same angular frequency $\omega$, given by $x = A_R\sin(\omega t + \phi_R)$, where $A_R$ is the resultant amplitude and $\phi_R$ is the resultant initial phase.

We can find the resultant amplitude by summing the phasors vectorially. Each phasor can be resolved into components along two perpendicular directions, for example, along the direction of $\sin \omega t$ (x-axis) and $\cos \omega t$ (y-axis). A displacement $A\sin(\omega t + \phi)$ can be written as $A(\sin \omega t \cos \phi + \cos \omega t \sin \phi) = (A\cos \phi)\sin \omega t + (A\sin \phi)\cos \omega t$. Let $S = A\cos \phi$ be the component along $\sin \omega t$ and $C = A\sin \phi$ be the component along $\cos \omega t$.

For $x_1 = 5\sin \omega t$: $A_1 = 5$, $\phi_1 = 0^\circ$. $S_1 = A_1 \cos \phi_1 = 5 \cos 0^\circ = 5 \times 1 = 5$. $C_1 = A_1 \sin \phi_1 = 5 \sin 0^\circ = 5 \times 0 = 0$.

For $x_2 = 5\sin(\omega t + 53^\circ)$: $A_2 = 5$, $\phi_2 = 53^\circ$. We use the approximate values for $\sin 53^\circ \approx 0.8$ and $\cos 53^\circ \approx 0.6$ (corresponding to a 3-4-5 right triangle). $S_2 = A_2 \cos \phi_2 = 5 \cos 53^\circ \approx 5 \times 0.6 = 3$. $C_2 = A_2 \sin \phi_2 = 5 \sin 53^\circ \approx 5 \times 0.8 = 4$.

For $x_3 = 10\sin(\omega t - 90^\circ)$: $A_3 = 10$, $\phi_3 = -90^\circ$. $S_3 = A_3 \cos \phi_3 = 10 \cos(-90^\circ) = 10 \times 0 = 0$. $C_3 = A_3 \sin \phi_3 = 10 \sin(-90^\circ) = 10 \times (-1) = -10$.

The resultant displacement is $x = (S_1 + S_2 + S_3)\sin \omega t + (C_1 + C_2 + C_3)\cos \omega t$. Let the resultant components be $S_R = S_1 + S_2 + S_3$ and $C_R = C_1 + C_2 + C_3$. $S_R = 5 + 3 + 0 = 8$. $C_R = 0 + 4 + (-10) = -6$.

The resultant displacement is $x = 8\sin \omega t - 6\cos \omega t$. This is in the form $x = S_R \sin \omega t + C_R \cos \omega t$. The resultant amplitude $A_R$ is given by $\sqrt{S_R^2 + C_R^2}$. $A_R = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.

Alternatively, using complex numbers: The complex amplitude for $A\sin(\omega t + \phi)$ is $A e^{i\phi} = A(\cos \phi + i\sin \phi)$. $Z_1 = 5 e^{i0^\circ} = 5(\cos 0^\circ + i\sin 0^\circ) = 5(1 + 0i) = 5$. $Z_2 = 5 e^{i53^\circ} = 5(\cos 53^\circ + i\sin 53^\circ) \approx 5(0.6 + i0.8) = 3 + 4i$. $Z_3 = 10 e^{i(-90^\circ)} = 10(\cos(-90^\circ) + i\sin(-90^\circ)) = 10(0 - i1) = -10i$.

The resultant complex amplitude is $Z_R = Z_1 + Z_2 + Z_3$. $Z_R = 5 + (3 + 4i) + (-10i) = (5 + 3) + (4i - 10i) = 8 - 6i$.

The resultant amplitude $A_R$ is the magnitude of the resultant complex amplitude $Z_R$. $A_R = |Z_R| = |8 - 6i| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.

Both methods yield the same resultant amplitude.

Explanation of the solution: The given SHMs are represented as phasors. The resultant displacement is the sum of the individual displacements. This sum corresponds to the vector sum of the individual phasors. We convert all SHM equations to the standard form $A\sin(\omega t + \phi)$. The phasor for $A\sin(\omega t + \phi)$ is a vector of length $A$ at an angle $\phi$ with respect to the reference axis. We resolve each phasor into components along the $\sin \omega t$ and $\cos \omega t$ directions (or equivalent perpendicular axes). Summing the components gives the components of the resultant phasor. The magnitude of the resultant phasor is the resultant amplitude, calculated using the Pythagorean theorem on the resultant components. Alternatively, using complex numbers, the resultant complex amplitude is the sum of individual complex amplitudes $A e^{i\phi}$. The magnitude of the resultant complex amplitude is the resultant amplitude. Both methods give the resultant amplitude as 10.