A bath toy is floating stationary in water such that half of the toy is above water and half of the toy is below the water. A kid pulls the toy 3.3 cm down into the water and then lets go. The toy then proceeds to bob up and down. It bobs up and down 18 times in a minute.
NOTE: Let up be positive and down be negative for this problem
(a) What is the Period of the motion?
(b) What is the Frequency of motion?
(c) What is the Amplitude of the motion?
(d) What is the equation that describes the position of the object at any given time?
(e) Where is the toy at time t = 7.92 second?
(f) Give meaning to your answer in part (e). Is the toy mostly above the water or below the water at this time. Explain with evidence how you know this.
(a) The Period of the motion is the time it takes for one complete cycle of motion. Since the toy bobs up and down 18 times in a minute, the period is 1 minute divided by 18, which is approximately 3.33 seconds.
(b) The Frequency of motion is the number of cycles per unit time. In this case, the frequency is 18 cycles per minute, or 0.3 cycles per second.
(c) The Amplitude of the motion is the maximum distance from the equilibrium position. In this case, the toy was pulled 3.3 cm down into the water, so the amplitude is 3.3 cm.
(d) The equation that describes the position of the object at any given time is: y(t) = A * sin(2π * f * t), where A is the amplitude, f is the frequency, and t is the time.
(e) To find where the toy is at t = 7.92 seconds, we substitute t = 7.92 seconds into the equation y(t) = 3.3 * sin(2π * 0.3 * 7.92). This gives us the position of the toy at that time.
(e) The toy will be mostly above the water at t = 7.92 seconds because the position is positive. This means the toy is above its equilibrium position.
(f) The evidence for this is the positive value obtained from the equation in part (e). Since the position is positive, the toy is mostly above the water at this time.