A spring is suspended from the ceiling. It is pulled down and released, and then oscillates up and down. Its length, \( l \) centimetres, is modelled by the function \[ l=33+5 \cos \left((720 t)^{\circ}\right) \] where \( t \) is time in seconds after release. (a) Find the length of the spring after 1 second. (b) Find (I) the minimum length of the spring. (ii) the maximum length of the spring. (c) Find the first time at which the length is 33 cm . (d) What is the period of the motion? (e) Within the first 5 seconds, find the number of times the spring has (i) its minimum length (ii) a length of 30 cm .

To solve the problems step by step, let's break them down. Given the function: \[ l = 33 + 5 \cos \left((720 t)^{\circ}\right) \] ### (a) Find the length of the spring after 1 second. Substituting \( t = 1 \) into the equation: \[ l = 33 + 5 \cos (720 \times 1^{\circ}) = 33 + 5 \cos (720^{\circ}) \] Since \( \cos(720^{\circ}) = \cos(0^{\circ}) = 1 \): \[ l = 33 + 5 \times 1 = 33 + 5 = 38 \text{ cm} \] ### (b) Find (i) the minimum length of the spring. (ii) the maximum length of the spring. (i) The maximum of \( \cos(x) \) is 1, so the maximum length is: \[ l_{\text{max}} = 33 + 5 \times 1 = 38 \text{ cm} \] The minimum of \( \cos(x) \) is -1, so the minimum length is: \[ l_{\text{min}} = 33 + 5 \times (-1) = 33 - 5 = 28 \text{ cm} \] ### (c) Find the first time at which the length is 33 cm. Setting \( l = 33 \): \[ 33 = 33 + 5 \cos \left((720 t)^{\circ}\right) \] \[ 0 = 5 \cos \left((720 t)^{\circ}\right) \] \[ \cos \left((720 t)^{\circ}\right) = 0 \] The cosine function is 0 at \( \frac{90}{720} + k \cdot \frac{180}{720} \) degrees, where \( k \) is an integer. For the first positive solution (\( k = 0 \)): \[ 720 t = 90 \implies t = \frac{90}{720} = \frac{1}{8} \text{ seconds} \] ### (d) What is the period of the motion? The angular frequency \( \omega \) in degrees is given by \( 720^{\circ} \). The period \( T \) in seconds can be calculated using the formula: \[ T = \frac{360^{\circ}}{720^{\circ}} \text{ seconds} \] This simplifies to: \[ T = \frac{1}{2} \text{ seconds} \] ### (e) Within the first 5 seconds, find the number of times the spring has (i) its minimum length (ii) a length of 30 cm. (i) The spring reaches its minimum length every half period, so in 5 seconds: \[ \text{Number of periods} = \frac{5 \text{ seconds}}{\frac{1}{2} \text{ seconds}} = 10 \] The minimum length occurs once per period, so it touches the minimum length 10 times. (ii) To find the times when \( l = 30 \): \[ 30 = 33 + 5 \cos((720 t)^{\circ}) \] \[ -3 = 5 \cos((720 t)^{\circ}) \] \[ \cos((720 t)^{\circ}) = -\frac{3}{5} \] The cosine is negative at angles of \( 180^{\circ} + x \) where \( x = \cos^{-1}\left(-\frac{3}{5}\right) \). We find \( x \approx 126.87^{\circ} \). The angles where this occurs in a full cycle (0 to 360 degrees) are: 1. \( (720 t)^{\circ} = 180^{\circ} + 126.87^{\circ} = 306.87^{\circ} \) 2. \( (720 t)^{\circ} = 180^{\circ} - 126.87^{\circ} = 53.13^{\circ} \) Thus we obtain: \[ t_1 = \frac{53.13}{720} , \quad t_2 = \frac{306.87}{720} \] Repeating these every full cycle gives approximately \( 1 \) time in each of the next \( 10 \) intervals. In total, this gives about \( 10 \) times. Therefore, in the first 5 seconds: - **Minimum length**: 10 times