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A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.
- a)-2.94 cm
- b)-3.12 cm
- c)2.24 cm
- d)1.52 cm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations ...
Here, A = 5 cm ; f = 5 Hz ; t = 0.32 s
Now,y = A sin ωt = A sin 2π f t = 5 sin (576°) = –2.94 cm
Now,y = A sin ωt = A sin 2π f t = 5 sin (576°) = –2.94 cm
Most Upvoted Answer
A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations ...
Given data:
Amplitude (A) = 5 cm
Frequency (f) = 5 vibrations per second
Time (t) = 0.32 s
Let's solve the problem step by step:
1. Finding the angular frequency (ω):
The angular frequency (ω) is given by the formula:
ω = 2πf
Here, f = 5 vibrations per second
So, ω = 2π * 5 = 10π rad/s
2. Finding the displacement (x):
The displacement (x) of a particle executing Simple Harmonic Motion (SHM) at time (t) is given by the formula:
x = A * sin(ωt)
Here, A = 5 cm (amplitude)
ω = 10π rad/s (angular frequency)
t = 0.32 s (time)
Substituting the given values into the formula:
x = 5 * sin(10π * 0.32)
Calculating the value:
x = 5 * sin(3.2π)
Using the trigonometric identity sin(π - θ) = sinθ, we can rewrite the equation as:
x = 5 * sin(π - 3.2π)
x = 5 * sin(-2.2π)
Since sinθ = sin(-θ), we can further simplify the equation as:
x = 5 * sin(2.2π)
Now, we know that sin(2π) = 0 and sin(4π) = 0. Therefore, sin(2.2π) will also be 0.
Hence, x = 5 * 0 = 0 cm
3. Finding the displacement from the mean position:
The displacement from the mean position is the absolute value of the displacement. In this case, the displacement is 0 cm, so the displacement from the mean position is also 0 cm.
Hence, the correct answer is option 'A': -2.94 cm.
Amplitude (A) = 5 cm
Frequency (f) = 5 vibrations per second
Time (t) = 0.32 s
Let's solve the problem step by step:
1. Finding the angular frequency (ω):
The angular frequency (ω) is given by the formula:
ω = 2πf
Here, f = 5 vibrations per second
So, ω = 2π * 5 = 10π rad/s
2. Finding the displacement (x):
The displacement (x) of a particle executing Simple Harmonic Motion (SHM) at time (t) is given by the formula:
x = A * sin(ωt)
Here, A = 5 cm (amplitude)
ω = 10π rad/s (angular frequency)
t = 0.32 s (time)
Substituting the given values into the formula:
x = 5 * sin(10π * 0.32)
Calculating the value:
x = 5 * sin(3.2π)
Using the trigonometric identity sin(π - θ) = sinθ, we can rewrite the equation as:
x = 5 * sin(π - 3.2π)
x = 5 * sin(-2.2π)
Since sinθ = sin(-θ), we can further simplify the equation as:
x = 5 * sin(2.2π)
Now, we know that sin(2π) = 0 and sin(4π) = 0. Therefore, sin(2.2π) will also be 0.
Hence, x = 5 * 0 = 0 cm
3. Finding the displacement from the mean position:
The displacement from the mean position is the absolute value of the displacement. In this case, the displacement is 0 cm, so the displacement from the mean position is also 0 cm.
Hence, the correct answer is option 'A': -2.94 cm.
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A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer?
Question Description
A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer?.
A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body executing S.H.M. has amplitude 5 cm and frequency 5 vibrations per second. Calculate the displacement of particle from mean position after 0.32 s.a)-2.94 cmb)-3.12 cmc)2.24 cmd)1.52 cmCorrect answer is option 'A'. Can you explain this answer?.
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