Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Notice that both strings have the same length, so they form an isosceles triangle, which means the angles they make with the ceiling are congruent.
If this angle has measure θ, then the angle between the two strings T₁ and T₂ has measure 180° - 2θ. This is because the interior angles of any triangle sum to 180°.
By the law of cosines, we have
(1 m)² = (0.87 m)² + (0.87 m)² - 2 (0.87 m)² cos(180° - 2θ)
Solve for θ :
(1 m)² = 2 (0.87 m)² - 2 (0.87 m)² cos(180° - 2θ)
(1 m)² / (2 (0.87 m)²) = 1 - cos(180° - 2θ)
cos(180° - 2θ) = 1 - (1 m)² / (2 (0.87 m)²)
cos(180° - 2θ) ≈ 0.3394
180° - 2θ = arccos(0.3394)
180° - 2θ ≈ 70.159°
2θ ≈ 109.841°
θ ≈ 54.9205° ≈ 55°
