Sound Waves Solved Example

SOUND WAVES Page # 22 SOLVED EXAMPLE Ex.1 On an average a human heart is found to beat Ex.3 Which of the following functions of time 75 times in a minute. Calculate its frequency and represent (a) simple harmonic motion and (b) period. periodic but not simple harmonic? Give the period Ans. The beat frequency of heart = 75/(1 min) = 75/(60 s) = 1.25 s –1 = 1.25 Hz The time period T = 1/(1.25 s–1) = 0.8 s Ex.2 Which of the following functions of time for each case. (1) sin t – cos t (2) sin2 t Ans. (a) sin t – cos t = sin t – sin (/2 – t) 2 cos (/4) sin (t – /4) = 2 sin (t – /4) represent (a) periodic and (b) non-periodic = motion? Give the period for each case of periodic This function represents a simple harmonic motion motion [ is any positive constant]. having a period T = 2/ and a phase angle (–/4) or (i) sin t + cos t (7/4) (b) sin2 t = ½ – ½ cos2 t (ii) sin t + cos 2 t + sin 4 t The function is periodic having a period T = /. It (iii) e–t also represents a harmonic motion with the point of (iv) log (t) equilibrium occurring at ½ instead of zero. Ans. (i) sin t + cos t is a periodic function, it can Ex.4 Figure depicts two circular motions. The also be written as 2 sin (t + /4). radius of the circle, the period of revolution, the Now 2 sin (t + /4)= = 2 sin (t + /4 + 2) 2 sin [(t + 2/) + /4] The periodic time of the function is 2/. (ii) This is initial position and the sense of revolution are indicated on the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its value, sin t has a period T0 = 2/; cos 2t has a period / =T0/2; and sin 4t has a period 2/4 = Ans. T0/4. The period of the first term is a multiple of the (a) At t = 0, OP makes an angle of 45º = /4 rad periods of the last two terms. Therefore, the smallest with the (positive direction of ) x-axis. After time t, it interval of time after which the sum of the three terms covers an angle t T 2in the anticlockwise sense, repeats is T0, and thus the sum is a periodic function with a period 2/. (iii) The function e–t is not periodic, it decreases monotonically with increasing time and tends to zero as t and thus, never repeats its value. and makes an angle of 2 t + /4 with the x-axis. T The projection of OP on the x-axis at time t (iv) The function log(t) increases monotonically with time t. It, therefore, never repeats its value and is a non-periodic function. It may be noted that as log(t) diverges to ?. It, therefore, cannot represent any kind of physical displacement. is given by, (t) = A cos ( 2 t + /4) for T = 4s, x(t) T = A cos (2/4 + /4) SOUND WAVES Page # 23 which is a SHM of amplitude A, period 4 s, and an Ex.6 Two identical springs of spring constant k are initial phase = /4 attached to a blockof mass m and to fixed supports (b) In this case at t = 0, OP makes an angle of with as shown in Fig. Show that when the mass is the x-axis. After a time t, it covers an angle of displaced from its equilibrium position on either 90º = 2/2 in the clockwise sense and makes an angle side, it executes a simple harmonic motion. Find 2 of t with the x-axis. The projection of OP on the T the period of oscillations. x-axis at time t is given by x(t) = B cos 2    t  Writing this as x (t ) = B cos   15 2   x(t) = A cos (t + 4) comparing with. We find that this represents a SHM of amplitude B, period 30 s, and an initial phase of   2 Ex.5 A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2t + /4]. At Ans. Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig. Under thissituation the spring on the left side gets t = 1.5 s, calculate the (a) displacement, (b) speed and (c) acceleration of the body. Ans.The angular frequency of the body = 2s–1 and its time period T = 1 s. At t = 1.5 s (a) displacement = (5.0 m) cos [(2s–1) × 1.5 s + /4] = (5.0 m) cos [(3 + /4)] = –5.0 × 0.707 m = –3.535 m (b) Using Eq. (14.9), the speed of the body = – (5.0 m)(2 s–1) sin [(2 s–1) ×1.5 s + /4] = – (5.0 m)(2 s–1) sin [(3 + /4)] = 10 × 0.707 m s–1 = 22 m s–1 (c) Using Eq., v(t) = d x(t) the acceleration of the dt body = –(2 s–1)2 × displacement = – (2 s–1)2 × (–3.535 m) = 140 m s–2 elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then, F1 = –k x (force exerted by the spring onthe left side, trying to pull the mass towards the mean position) F2 = –k x (force exerted by the spring on the right side, trying to push the mass towards the mean position) The net force, F, acting on the mass is then given by, F = –2kx Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is, T= m 2k