A string of length 20 cm and linear mass density 0.40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t=0 near an end as shown in figure, which travels towards the other end. After what time in milliseconds, the string will have same shape (shown in diagram) again ?

Question

A string of length 20 cm and linear mass density 0.40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t=0 near an end as shown in figure, which travels towards the other end. After what time in milliseconds, the string will have same shape (shown in diagram) again ? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-lhzfoa4setsibnbr.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given

μ =0.40 g/cm=(0 . 40/1000)× 100=0.040 kg/m Wave speed in the stretched string v=√(F/μ )=√(16/0 . 040)=20 m/s The string will regain its shape after travelling the pulse a distance =20+20=40 cm Thus time speed t=(0 . 40/20)=0.02 s

Given μ=0.40g/cm=10000.40​×100=0.040kg/m Wave speed in the stretched string v=μF​​=0.04016​​=20m/s The string will regain its shape after travelling the pulse a distance =20+20=40cm Thus time speed t=200.40​=0.02s

Given
$μ = 0.40 g / c m = \frac{0.40}{1000} \times 100 = 0.040 k g / m$
Wave speed in the stretched string
$v = \frac{F}{μ} = \frac{16}{0.040} = 20 m / s$
The string will regain its shape after travelling the pulse a distance
$= 20 + 20 = 40 c m$
Thus time speed $t = \frac{0.40}{20} = 0.02 s$