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A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be?
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A particle performing SHM with a frequency 10 hearts and amplitude 5 c...
X = 0.05 sin ( 20 π t + π )
In general, x = A sin ( ωt + ϕ )
Amplitude A = 5 c m
frequency f = 10
⇒ ω = 2π f
ω = 20π
⇒ x = 5 cm sin (20 πt + ϕ)
Since x = − 5 cm at t = 0
− 5 = 5 sin ϕ
⇒ ϕ = π
∴ x = 5 cm sin (20 πt + π)
As x is in meters
∴ x = 0.05 sin (20 πt + π)
In general, x = A sin ( ωt + ϕ )
Amplitude A = 5 c m
frequency f = 10
⇒ ω = 2π f
ω = 20π
⇒ x = 5 cm sin (20 πt + ϕ)
Since x = − 5 cm at t = 0
− 5 = 5 sin ϕ
⇒ ϕ = π
∴ x = 5 cm sin (20 πt + π)
As x is in meters
∴ x = 0.05 sin (20 πt + π)
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A particle performing SHM with a frequency 10 hearts and amplitude 5 c...
Explanation of SHM with given parameters
Defining SHM
Simple Harmonic Motion (SHM) is the motion of an object back and forth around its mean position under the influence of a restoring force that is directly proportional to the displacement from the mean position.
Frequency and Amplitude
- The frequency of SHM is the number of oscillations per second and is denoted by 'f'.
- Here, the frequency is given as 10 Hz.
- The amplitude of SHM is the maximum displacement of the particle from its mean position and is denoted by 'A'.
- Here, the amplitude is given as 5 cm.
Displacement of the particle
The displacement of the particle at any time 't' can be given by the equation:
x = A cos(2πft)
where,
x = displacement of the particle
A = amplitude of SHM
f = frequency of SHM
t = time
In this case, the particle is initially in left extreme position, which means its displacement is maximum and negative. Therefore, we can write:
x = -5 cos(2π10t)
Visualization of displacement
As the particle is performing SHM, its displacement will keep changing with time. The displacement-time graph of the particle will be a sinusoidal wave with a frequency of 10 Hz and an amplitude of 5 cm. The particle will move back and forth around its mean position, crossing the mean position twice in each cycle.
Conclusion
In conclusion, the displacement of the particle performing SHM with a frequency of 10 Hz and an amplitude of 5 cm initially in left extreme position can be given by the equation x = -5 cos(2π10t). The particle will move back and forth around its mean position, crossing the mean position twice in each cycle.
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A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be?
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A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be?.
A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle performing SHM with a frequency 10 hearts and amplitude 5 centimetre initially in left extreme position vision of displacement will be?.
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