A 40.1 g object is attached to a horizontal spring with a spring constant of 11.9 N/m and released from rest with an amplitude o
1 answer:
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Answer:
When both initial speed and initial displacement is doubled then amplitude will be doubled.
Explanation:
Given that :- Amplitude of simple harmonic Oscillator is doubled.
So,
Formula of Simple harmonic oscillator is ...........(1)
Where X = Position in (m,cm,km.....)
A = Amplitude in (m,cm,km.....)
F = Frequency in (Hz)
T = Time in (sec.)
Ф = Phase in (rad)
For initial displacement taking t=0 we get,
Initial displacement = .................(2)
Now taking equation (1) and differentiating it w.r.t to (t) we get
taking t=0 for initial speed then we get,
Initial speed = ...............(3)
observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.
Hence,
when both initial speed and displacement is doubled then amplitude will be doubled.
Answer:
1.01 × 10⁵ Pa
Explanation:
At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.
We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.
Pressure at a depth is given by :
Where
is the density of air,
So,
Total pressure, P = Atmospheric pressure + 12 Pa
= 1.013 × 10⁵ Pa + 12 Pa
= 1.01 × 10⁵ Pa
Hence, the total pressure is 1.01 × 10⁵ Pa.