Question

 

A transverse wave on a string is given by the formula D(x,t) = 0.15 sin(4.0x - 43.0t). where D and z are in m and t is in s 

Part A

At t = 7.0 x 10 2s. what is the transverse displacement of the point on the string where z-0.45m?

 Express your answer using two significant figures. 

D(0.45m, 7.0 × 10-2S) = 

Part B 

At t = 7.0 × 10 2s. what is the (transverse) velocity of the point on the string where x = 0.45m? 

Answer 1

(A)

Displacememt on the string is

D(0.5m , 7*10^-2 s) = 0.1*sin(3.6(0.5) - 41(7.4*10^-2) )

= 0.1sin(-1.234)

= 0.1*-0.0215

= -0.022 m

(B)

the velocity of the string is ,

$\delta$ D (x,t) / $\delta$ t = 0.10(-41)cos(3.6*x - 41*t)

$\delta$ D (0.45 , 7*10^-2) / $\delta$ t = -4.1 cos(3.6*0.45 - 4.1*7*10^-2)

= -4.1cos(1.62 - 0.287)

= -4.09 m/s

Answer 2

43モ = -45 m Sın ( r. 8-3-o↓ D , IS = -- 、14 3 m d D- -6.45 S ( i-8-3.01) - 2 2 s mls