A simple pendulum in a certain planet is found to vibrate 50 times in 100s. Find its (a) period, (b) its frequency. (c) When 1.5m of its length is cut, it vibrates 50 times in 50 s. Find the original length of the pendulum.
thanks in advance
(a) To find the period of the simple pendulum, we divide the time taken for one complete oscillation by the number of oscillations.
Given:
Number of oscillations = 50
Time taken for 50 oscillations = 100s
The period, denoted as T, is given by:
T = (Time taken for 50 oscillations) / (Number of oscillations)
T = 100s / 50 = 2s
Therefore, the period of the simple pendulum is 2 seconds.
(b) The frequency, denoted as f, is the reciprocal of the period.
f = 1 / T
f = 1 / 2s = 0.5 Hz
Therefore, the frequency of the simple pendulum is 0.5 Hz.
(c) Let's assume the original length of the pendulum is L.
Given:
Number of oscillations = 50
Time taken for 50 oscillations = 50s
Length of the pendulum after cutting = L - 1.5m
The period of the pendulum is proportional to the square root of the length.
Therefore, (Time taken for 50 oscillations) / (Number of oscillations) = K * √(Length of the pendulum)
where K is a constant.
Plugging in the given values, we have:
50s / 50 = K * √(L - 1.5m)
1s = K * √(L - 1.5m)
Squaring both sides of the equation, we get:
1s^2 = K^2 * (L - 1.5m)
1 = K^2 * (L - 1.5m)
Since K^2 is also a constant, we can write the equation as:
1 = Constant * (L - 1.5m)
1 = Constant * L - Constant * 1.5m
1 + Constant * 1.5m = Constant * L
Now, we know that when 1.5m of its length is cut, it vibrates 50 times in 50s. Therefore, we can set up another equation:
50s / 50 = Constant * √(L - 1.5m)
1s = Constant * √(L - 1.5m)
Squaring both sides of the equation, we get:
1s^2 = Constant^2 * (L - 1.5m)
1 = Constant^2 * (L - 1.5m)
Since Constant^2 is also a constant, we can write the equation as:
1 = Constant * L - Constant * 1.5m
1 + Constant * 1.5m = Constant * L
Comparing the two equations, we have:
1 + Constant * 1.5m = Constant * L
1 + Constant * 1.5m = 1 + Constant * 1.5m
Both equations are the same, meaning that the original length of the pendulum is not affected by the length cut.
Therefore, the original length of the pendulum is L.
To find the answers to the questions, we need to understand the definitions of period and frequency for a simple pendulum.
(a) The period of a pendulum is defined as the time it takes for the pendulum to complete one full oscillation (i.e., one back-and-forth motion).
(b) The frequency of a pendulum is defined as the number of oscillations it completes in a given time interval.
Now, let's solve the given problem step by step:
(a) To find the period, we need to divide the total time by the number of oscillations. In this case, the pendulum vibrates 50 times in 100 seconds. Therefore, the period (T) can be calculated as T = total time / number of oscillations = 100 seconds / 50 = 2 seconds.
(b) To find the frequency, we need to divide the number of oscillations by the total time. In this case, the pendulum vibrates 50 times in 100 seconds. Therefore, the frequency (f) can be calculated as f = number of oscillations / total time = 50 / 100 = 0.5 Hz.
Now, let's move on to the second part of the question:
(c) When 1.5m of the pendulum's length is cut, the new length of the pendulum becomes unknown. However, we know that it vibrates 50 times in 50 seconds.
To solve for the original length of the pendulum, we can use the following relationship:
(T1 / T2) = √(L1 / L2)
where T1 and T2 are the periods of the pendulum for the initial and final lengths, and L1 and L2 are the initial and final lengths of the pendulum.
Let's assign some variables:
T1 = initial period (unknown)
T2 = 50 seconds
L1 = initial length (unknown)
L2 = L1 - 1.5m (new length after cutting)
Since we know T2 = 50 seconds, we can substitute the values into the equation:
(2 seconds / 50 seconds) = √(L1 / (L1 - 1.5 meters))
By squaring both sides of the equation, we get:
(2/50)^2 = L1 / (L1 - 1.5)
(4/2500) = L1 / (L1 - 1.5)
Multiplying both sides by (L1 - 1.5), we get:
(4/2500) * (L1 - 1.5) = L1
Rearranging the equation, we have:
(4/2500) * L1 - (4/2500) * 1.5 = L1
Multiplying through by 2500 to remove the denominators, we get:
4L1 - 6 = 2500L1
3L1 = 6
L1 = 2 meters
Therefore, the original length of the pendulum is 2 meters.
In summary, the answers to the questions are:
(a) The period of the pendulum is 2 seconds.
(b) The frequency of the pendulum is 0.5 Hz.
(c) The original length of the pendulum is 2 meters.
(a) The original pendulum has a period of
P = 2.0 seconds.
(b) f = 1/P = 0.5 Hz
(c) Cutting off 1.5 meters reduces the period to 1.0 s. The original length L was four times the final length L', since Period is proportional so sqrt L.
L' = L - 1.5 = 0.25 L
L = 0.75 L = 1.5
L = 2.0 m