At 150 degrees, what is the amplitude of a sine wave with a peak value of 5 volts?

At 150 degrees, what is the amplitude of a sine wave with a peak value of 5 volts?

• A. +2.5 volts
• B. +1.5 volts
• C. -2.5 volts
• D. 0 volts

Answer: (A)

To determine the amplitude of a sine wave at a specific angle, it's important to understand that the amplitude represents the maximum value of the wave. The peak value is the highest point the sine wave reaches above and below the zero line. The formula for the sine wave can be represented as: \[ V(t) = V_{peak} \times \sin(\theta) \] where \( V_{peak} \) is the peak amplitude (5 volts in this case) and \( \theta \) is the angle in degrees. At 150 degrees, we can calculate the sine value: \[ \sin(150^\circ) = 0.5 \] Now, we substitute this value back into the equation: \[ V(t) = 5 \text{ volts} \times 0.5 = 2.5 \text{ volts} \] The sine value is positive in this quadrant, so the amplitude at 150 degrees is +2.5 volts. Therefore, the answer reflects the sine wave's value at that specific angle, which correctly indicates the wave is above the zero line at that point, leading to the conclusion that the amplitude is +2.5 volts.

To determine the amplitude of a sine wave at a specific angle, it's important to understand that the amplitude represents the maximum value of the wave. The peak value is the highest point the sine wave reaches above and below the zero line.

The formula for the sine wave can be represented as:

[ V(t) = V_{peak} \times \sin(\theta) ]

where ( V_{peak} ) is the peak amplitude (5 volts in this case) and ( \theta ) is the angle in degrees. At 150 degrees, we can calculate the sine value:

[ \sin(150^\circ) = 0.5 ]

Now, we substitute this value back into the equation:

[ V(t) = 5 \text{ volts} \times 0.5 = 2.5 \text{ volts} ]

The sine value is positive in this quadrant, so the amplitude at 150 degrees is +2.5 volts. Therefore, the answer reflects the sine wave's value at that specific angle, which correctly indicates the wave is above the zero line at that point, leading to the conclusion that the amplitude is +2.5 volts.