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Two simple pendulums of length 1m and 25 m, respectively, are both given small displacements in the same direction at the same instant.If they are in phase after the shorter pendulum has completed n oscillation, n is equal to
- a)4/3
- b)5/4
- c)7/5
- d)8/7
Correct answer is option 'B'. Can you explain this answer?
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Two simple pendulums of length 1m and 25 m, respectively, are both giv...
∴ T ∝ √L , as time period decreases when the length of pendulum decreases, the time period of shorter pendulum (Ts) is smaller than that of longer pendulum (Tl). That means shorter pendulum performs more oscillations in a given time.
It is given that after n oscillations of shorter pendulum, both are again in phase. So, by this time longer pendulum must have made (n – 1) oscillations.
So, the two pendulum shall be in the same phase for the first time when the shorter pendulum has
completed 5/4 oscillation
completed 5/4 oscillation
Most Upvoted Answer
Two simple pendulums of length 1m and 25 m, respectively, are both giv...
Explanation:
To understand why the answer is option 'B', let's break down the problem step by step.
1. Period of a Simple Pendulum:
The period of a simple pendulum is given by the formula:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
2. Relationship between Period and Frequency:
The frequency of a pendulum is the reciprocal of its period:
f = 1/T
3. Phase Difference:
When two pendulums are in phase, it means they reach their maximum displacement in the same direction at the same time.
4. Relationship between Frequency and Phase Difference:
The phase difference between two pendulums can be determined by comparing their frequencies. If two pendulums have frequencies f1 and f2, the number of oscillations completed by the first pendulum when they are in phase is given by:
n = (f2/f1) = (T1/T2)
5. Applying the Concept:
In this problem, we are given two pendulums with lengths 1m and 25m respectively. Let's denote the period of the shorter pendulum as T1 and the period of the longer pendulum as T2.
From the formula for the period of a simple pendulum, we have:
T1 = 2π√(1/g) and T2 = 2π√(25/g)
Taking the ratio of the periods:
T2/T1 = (√(25/g))/(√(1/g)) = √(25/1) = 5
Since the ratio of the periods is 5, the ratio of the frequencies is also 5.
Therefore, the number of oscillations completed by the shorter pendulum when they are in phase is:
n = (T2/T1) = 5
Hence, the correct answer is option 'B': 5/4.
To understand why the answer is option 'B', let's break down the problem step by step.
1. Period of a Simple Pendulum:
The period of a simple pendulum is given by the formula:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
2. Relationship between Period and Frequency:
The frequency of a pendulum is the reciprocal of its period:
f = 1/T
3. Phase Difference:
When two pendulums are in phase, it means they reach their maximum displacement in the same direction at the same time.
4. Relationship between Frequency and Phase Difference:
The phase difference between two pendulums can be determined by comparing their frequencies. If two pendulums have frequencies f1 and f2, the number of oscillations completed by the first pendulum when they are in phase is given by:
n = (f2/f1) = (T1/T2)
5. Applying the Concept:
In this problem, we are given two pendulums with lengths 1m and 25m respectively. Let's denote the period of the shorter pendulum as T1 and the period of the longer pendulum as T2.
From the formula for the period of a simple pendulum, we have:
T1 = 2π√(1/g) and T2 = 2π√(25/g)
Taking the ratio of the periods:
T2/T1 = (√(25/g))/(√(1/g)) = √(25/1) = 5
Since the ratio of the periods is 5, the ratio of the frequencies is also 5.
Therefore, the number of oscillations completed by the shorter pendulum when they are in phase is:
n = (T2/T1) = 5
Hence, the correct answer is option 'B': 5/4.
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Two simple pendulums of length 1m and 25 m, respectively, are both given small displacements in the same direction at the same instant.If they are in phase after the shorterpendulum has completed n oscillation, nis equal toa)4/3b)5/4c)7/5d)8/7Correct answer is option 'B'. Can you explain this answer?
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