A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of $160 N m^{- 1}$ and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring ?

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Solution

Given:
$l = 40 c m, m a s s = 10 g$
$∴$ mass per unit length, $m = 10 / 40 = 1 / 4 (g / c m)$
spring constant $K = 160 N / m$
deflection $= x = 1 c m = 0.01 m$
$\Rightarrow T = k x = 160 \times 0.01 = 1.6 N = 16 \times 10^{4} d y n e$
Again $v = \sqrt{(T / m)} = \sqrt{(16 \times 10^{4} / (1 / 4)} = 8 \times 10^{2} c m / s = 800 c m / s$
$∴$ Time taken by the pulse to reach the spring
$t = 40 / 800 = 1 / 20 = 0 / 05 s e c .$

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A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of 160Nm−1 and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring ?

Given:l=40cm,mass=10g∴ mass per unit length, m=10/40=1/4(g/cm)spring constant K=160N/mdeflection =x=1cm=0.01m⇒T=kx=160×0.01=1.6N=16×104dyneAgain v=√(T/m)=√(16×104/(1/4)=8×102cm/s=800cm/s∴ Time taken by the pulse to reach the springt=40/800=1/20=0/05sec.