Most Important Selected Qs for JEE AdvancedPhysicsWaves and Sound
A pulse is started at a time t=0 along the +x direction on a long, taut string. The shape of the pulse at t=0 is given by function f(x) with f(x)= array ccc x 4 +1 & for & -4 x 0 -x+1 & for & 0 x 1 0 & & otherwise array . here f and x are in centimeters. The linear mass density of the string is 50 ~g / m and it is under a tension of 5 N .
Options
- AThe shape of the string is drawn at t=0 then the area of the pulse enclosed by the string and the x -axis is 2
- BThe shape of the string is drawn at t =0 then the area of the pulse enclosed by the string and the x -axis is
- CThe transverse velocity of the particle at x=13 ~cm and t =0.015 ~s will be -250 ~cm / s
- DThe transverse velocity of the particle at x =13 ~cm and t =0.015 ~s will be 250 ~cm / s
Correct answer
C. The transverse velocity of the particle at x=13 ~cm and t =0.015 ~s will be -250 ~cm / s
Step-by-step solution
The shape of the string will be Area = 1 2 5 1=2.5 ~cm ^2 Wave velocity =5 5 1000 50 ~m / s =10 ~m / s Thus the part with slope 1 4 will be present at x=13 & t=0.015 . v_P= 1 4 d x d t = -1 4 v_ wave =-250 ~cm / sec