Chapter 2: Q85P (page 58)

A man drops a rock into a well. (a) The man hears the sound of the splash 2.40 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water? (b) What If? If the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?

Short Answer

Expert verified

(a) The distance of water surface in well from top of well is 26.4m.

(b) The percentage error in distance from top to water surface in well due to ignore sound travel is 6.82%.

Step by step solution

Identification of given data

$T h e t i m e t o h e a r s o u n d o f s p l a s h a f t e r r e l e a s e o f r o c k i s t_{1} + t_{2} = 2.40 s . T h e s p e e d o f s o u n d i n a i r i s v = 336 m / s .$

Determination of distance of surface of water in well from top of well

$(a) T h e t i m e f o r t h e f a l l o f r o c k i n w e l l i s g i v e n a s : d = \frac{1}{2} g {t_{1}}^{2} . . . . . . (1) T h e t i m e f o r t h e s o u n d t o r e a c h a t t h e t o p o f w e l l i s g i v e n a s : d = v t_{2} . . . . . . . (2) C o m p a r e e q u a t i o n (1) a n d e q u a t i o n (2) . \frac{1}{2} g {t_{1}}^{2} = v t_{2} S u b s t i t u t e t_{1} = 2.40 s - t_{1} i n t h e a b o v e e q u a t i o n . \frac{1}{2} (9.8 m / s^{2}) (2.40 s - t_{1}) = (336 m / s) t_{2} {t_{2}}^{2} - (73.37) t_{2} + 5.76 t_{2} = 0.0786 s S u b s t i t u t e a l l t h e v a l u e s i n e q u a t i o n (2) . d = (336 m / s) (0.0786 s) d = 26.4 m T h e r e f o r e, t h e d i s \tan c e o f w a t e r s u r f a c e i n w e l l f r o m t o p o f w e l l i s 26.4 m .$

Determination of error in distance from top of well to water surface without sound travel

$T h e d i s \tan c e o f t o p o f w e l l f r o m w a t e r s u r f a c e, b y i g n o r i n g s o u n d t r a v e l, i s g i v e n a s : D = \frac{1}{2} g {(t_{1} + t_{2})}^{2} S u b s t i t u t e a l l t h e v a l u e s i n a b o v e e q u a t i o n . D = \frac{1}{2} (9.8 m / s^{2}) {(2.40 s)}^{2} D = 28.2 m T h e p e r c e n t a g e e r r o r i n d i s \tan c e f r o m t o p t o w a t e r s u r f a c e i n w e l l i s g i v e n a s : x = (\frac{D - d}{d}) \times 100 x = (\frac{28.2 m - 26.4 m}{26.4}) \times 100 x = 6.82 % T h e r e f o r e, t h e p e r c e n t a g e e r r o r i n d i s \tan c e f r o m t o p t o w a t e r s u r f a c e i n w e l l d u e t o i g n o r e i g n o r i n g o f s o u n d t r a v e l i s 6.82 % .$