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Tempoh ayunan, \( T \) suatu spring berpemberat berubah secara langsung dengan punca kuasa dua jisim pemberatnya, \( m \). Tempoh ayunan ialah 2.24 s apabila jisimnya 20 g . Cari The ascillating period, Ta loaded spring varies directly with square root of its mass of load, \( m \). The oscillating period is 2.24 s the mass is 20 g . Find (a) tempoh ayunan jika jisim pemberat ialah 64 g , ARAB \( \mathbf{S} \) the period of the oscillation if the load mass is 64 g , (b) jisim pemberat jika tempoh ayunannya ialah 3.16 s . ARAS \( \mathbf{8} \) the mass of the load if the period of oscillation is 3.16 s . [8 markah/,

Ask by Burton Vaughan. in Malaysia
Feb 18,2025

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Answer

(a) The period of oscillation when the mass is 64 g is approximately 4.01 s. (b) The mass of the load when the period of oscillation is 3.16 s is approximately 39.7 g.

Solution

To solve the problem, we will use the relationship given in the problem statement, which states that the oscillating period \( T \) varies directly with the square root of the mass \( m \). This can be expressed mathematically as: \[ T = k \sqrt{m} \] where \( k \) is a constant of proportionality. ### Step 1: Find the constant \( k \) We know that when \( m = 20 \) g, \( T = 2.24 \) s. We can substitute these values into the equation to find \( k \): \[ 2.24 = k \sqrt{20} \] Calculating \( \sqrt{20} \): \[ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \approx 4.472 \] Now substituting this back into the equation: \[ 2.24 = k \times 4.472 \] To find \( k \): \[ k = \frac{2.24}{4.472} \approx 0.501 \] ### Step 2: Solve for (a) the period of oscillation when the mass is 64 g Now we will use the value of \( k \) to find the period \( T \) when \( m = 64 \) g: \[ T = k \sqrt{64} \] Calculating \( \sqrt{64} \): \[ \sqrt{64} = 8 \] Now substituting \( k \): \[ T = 0.501 \times 8 \approx 4.008 \] ### Step 3: Solve for (b) the mass of the load when the period of oscillation is 3.16 s We will use the same equation to find \( m \) when \( T = 3.16 \) s: \[ 3.16 = k \sqrt{m} \] Substituting \( k \): \[ 3.16 = 0.501 \sqrt{m} \] To find \( \sqrt{m} \): \[ \sqrt{m} = \frac{3.16}{0.501} \approx 6.303 \] Now squaring both sides to find \( m \): \[ m = (6.303)^2 \approx 39.7 \] ### Final Answers (a) The period of oscillation when the mass is 64 g is approximately \( 4.01 \) s. (b) The mass of the load when the period of oscillation is 3.16 s is approximately \( 39.7 \) g.

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Untuk menyelesaikan bahagian (a), kita boleh menggunakan formula yang berkaitan dengan tempoh ayunan dan jisim pemberat: \[ T = k \sqrt{m} \] di mana \( k \) adalah pemalar. Dari data yang diberi, kita perlu mencari nilai \( k \) terlebih dahulu menggunakan nilai tempoh dan jisim yang ada. Oleh itu, kita ada: \[ 2.24 = k \sqrt{20} \] Setelah membahagikan kedua-duanya, kita boleh cari \( k \) dan seterusnya menggunakan nilai \( k \) untuk mencari \( T \) apabila jisim adalah 64 g. Untuk bahagian (b), kita akan menggunakan formula yang sama, tetapi kali ini kita tahu tempoh \( T \) dan mencari jisim \( m \): \[ 3.16 = k \sqrt{m} \] Dengan menggunakan nilai \( k \) daripada bahagian (a), kita boleh mencari nilai \( m \) secara langsung. Kira-kira dan dapatkan jawapan yang tepat!

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