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The phase difference between two SHM y(1...

The phase difference between two SHM `y_(1) = 10 sin (10 pi t + (pi)/(3))` and `y_(2) = 12 sin (8 pi t + (pi)/(4))` at `t = 0.5s` is

A

`(11 pi)/(12)`

B

`(13 pi)/(12)`

C

`pi`

D

`(17 pi)/(12)`

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To find the phase difference between the two simple harmonic motions (SHM) given by the equations \( y_1 = 10 \sin(10 \pi t + \frac{\pi}{3}) \) and \( y_2 = 12 \sin(8 \pi t + \frac{\pi}{4}) \) at \( t = 0.5 \) seconds, we will follow these steps: ### Step 1: Identify the Phase of Each SHM The phase of the first SHM \( y_1 \) is given by: \[ \phi_1 = 10 \pi t + \frac{\pi}{3} \] The phase of the second SHM \( y_2 \) is given by: \[ \phi_2 = 8 \pi t + \frac{\pi}{4} \] ### Step 2: Substitute the Given Time \( t = 0.5 \) seconds Now, we will substitute \( t = 0.5 \) into both phases. For \( y_1 \): \[ \phi_1 = 10 \pi (0.5) + \frac{\pi}{3} = 5\pi + \frac{\pi}{3} \] For \( y_2 \): \[ \phi_2 = 8 \pi (0.5) + \frac{\pi}{4} = 4\pi + \frac{\pi}{4} \] ### Step 3: Simplify the Phases Now we will simplify both expressions. For \( \phi_1 \): \[ \phi_1 = 5\pi + \frac{\pi}{3} = \frac{15\pi}{3} + \frac{\pi}{3} = \frac{16\pi}{3} \] For \( \phi_2 \): \[ \phi_2 = 4\pi + \frac{\pi}{4} = \frac{16\pi}{4} + \frac{\pi}{4} = \frac{17\pi}{4} \] ### Step 4: Calculate the Phase Difference The phase difference \( \Delta \phi \) is given by: \[ \Delta \phi = \phi_1 - \phi_2 \] Substituting the values we calculated: \[ \Delta \phi = \frac{16\pi}{3} - \frac{17\pi}{4} \] ### Step 5: Find a Common Denominator and Simplify To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12. Convert \( \frac{16\pi}{3} \) and \( \frac{17\pi}{4} \) to have a denominator of 12: \[ \frac{16\pi}{3} = \frac{64\pi}{12} \] \[ \frac{17\pi}{4} = \frac{51\pi}{12} \] Now, substituting back into the phase difference: \[ \Delta \phi = \frac{64\pi}{12} - \frac{51\pi}{12} = \frac{13\pi}{12} \] ### Final Answer Thus, the phase difference between the two SHMs at \( t = 0.5 \) seconds is: \[ \Delta \phi = \frac{13\pi}{12} \] ---

To find the phase difference between the two simple harmonic motions (SHM) given by the equations \( y_1 = 10 \sin(10 \pi t + \frac{\pi}{3}) \) and \( y_2 = 12 \sin(8 \pi t + \frac{\pi}{4}) \) at \( t = 0.5 \) seconds, we will follow these steps: ### Step 1: Identify the Phase of Each SHM The phase of the first SHM \( y_1 \) is given by: \[ \phi_1 = 10 \pi t + \frac{\pi}{3} \] The phase of the second SHM \( y_2 \) is given by: ...
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