Một chất điểm dao động điều hòa trên trục Ox. Khi chất điểm đi qua vị trí cân bằng thì tốc độ của nó là 20 cm/s. Khi chất điểm có tốc độ là 10 cm/s thì gia tốc của nó có độ lớn là . Biên độ dao động của chất điểm là
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- Vận tốc cực đại: $v_{max} = A\omega = 20$ cm/s (1)
- Khi $v = 10$ cm/s thì $a = 40\sqrt{3}$ cm/s²
Suy ra:
$a^2/\omega^4 + v^2/\omega^2 = A^2$
$(40\sqrt{3})^2 / \omega^4 + 10^2 / \omega^2 = A^2$ (2)
Từ (1) suy ra $\omega = 20/A$. Thay vào (2), ta được:
$(40\sqrt{3})^2 / (20/A)^4 + 10^2 / (20/A)^2 = A^2$
$((40\sqrt{3})^2 * A^4) / 20^4 + (10^2 * A^2) / 20^2 = A^2$
$((1600 * 3) * A^4) / 160000 + (100 * A^2) / 400 = A^2$
$(3A^4) / 100 + A^2 / 4 = A^2$
$(3A^4) / 100 = (3A^2) / 4$
$A^2 = 100/4$
$A^2 = 25$
$A = 5$ cm.
Vậy biên độ dao động của chất điểm là 5 cm.
Câu hỏi liên quan
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Mạch gỗ vận chuyển
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