If the ticker-tape timer has a frequency of 50 Hz, it means it punches 50 holes in 1 second.
Hence the time interval to punch 1 hole is 1/50 = 0.02s. Alternatively, you can use the formula Period, T = 1/f = 1/50 = 0.02 s.
The diagram shows a strip of paper tape that has been pulled under a vibrating arm by a car moving to the left. The arm is vibrating regularly, making 50 dots per second. What was the acceleration of the car?
Solutions: View the video tutorial to understand where to take time interval.
The arm vibrating 50 dots per second = frequency of 50 Hz (50 holes are produced in 1 second)
Hence, the period, T = 1/f = 1 / 50 = 0.02s (every 0.02s, a hole is created on the tape)
To find acceleration, a, we need to find initial velocity, u,and the final velocity, v.
u = dist / time = 0.02m/0.02s = 1 m/s
v = dist / time = 0.04m/0.02s = 2 m/s
To find acceleration, we can use a = (v-u)/t , but the time, t, taken for the increase in velocity is where most students will make a mistake.
Many will take 7 intervals to calculate the t, which is wrong.
But the intervals should be 6, starting from the middle of first interval and middle of last interval.
This will give a more accurate acceleration.
Hence, t = 6 x 0.02s = 0.12s a = (v-u)/t = (2-1)/0.12 = 8.33 m/s2
Many are familiar with ticker-tape timer. The tape is attached to Car B. As Car B moves to the left, it pulls the tape with it. The tape will go through the ticker-tape timer machine. The machine is stationary and it just punches holes on the tape as the tape is pulled to the left. It is clear that Car A is accelerating as the distance interval of the holes is getting wider.
For Car B, the scenario is different. Oil is leaking from the car onto the road at fixed interval or frequency. Hence Car B is decelerating as the distance interval of oil drops becomes smaller.
Therefore, be careful with the two scenarios and do not be confused.