A particle is performing shm .its path length is 8cm and time period is 0.2 sec. if the particle is initially in the mean position .find the time at which it will be at 1 cm and the phase angle
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Path length is 8 cm.
∴ Circumference of the entire circle in the figure = 8 cm
Which means 2∏ = 8 cm
∴ ∏ = 4 cm
∴∏/2 = 2 cm.
So ∏/4 = 1cm.
Which means at 45° from mean position (point A), it will be at a distance 1 cm.
So phase angle is 45°.
Now to find time at 1 cm i.e. at 45°:
45° = T/360 × 45
= T/8
So time taken is T/8.
Hope it helps. Thank you. ☺♥
∴ Circumference of the entire circle in the figure = 8 cm
Which means 2∏ = 8 cm
∴ ∏ = 4 cm
∴∏/2 = 2 cm.
So ∏/4 = 1cm.
Which means at 45° from mean position (point A), it will be at a distance 1 cm.
So phase angle is 45°.
Now to find time at 1 cm i.e. at 45°:
45° = T/360 × 45
= T/8
So time taken is T/8.
Hope it helps. Thank you. ☺♥
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