A body performs SHM along the straight line segment A B C D E with C as the mid point of segment A E(A and E are the extreme position for the SHM). Its kinetic energies at B and D are each one fourth of its maximum value. If length of segment A E is 2 R, then the distance between B and D is

Question

A body performs SHM along the straight line segment A B C D E with C as the mid point of segment A E(A and E are the extreme position for the SHM). Its kinetic energies at B and D are each one fourth of its maximum value. If length of segment A E is 2 R, then the distance between B and D is (A) (√3/2) R (B) (R/√2) (C) √3 R (D) √2 R

Tardigrade Admin · Accepted Answer

When the particle crosses D, its speed is half the maximum speed. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dd2e80c305a72a8899c23c6a1f698626-.png" /> ∴ V=(v max /R) √R2-x2 or (v max /2)=(v max /R) √R2-x2 or x=(√3/2) R Distance B D=2 x=√3 R

$\frac{3}{2} R$

$\frac{R}{2}$

$3 R$

$2 R$

When the particle crosses $D$ , its speed is half the maximum speed.

$∴ V = \frac{v _{m a x}}{R} R^{2} - x^{2}$
or $\frac{v _{m a x}}{2} = \frac{v _{m a x}}{R} R^{2} - x^{2}$ or $x = \frac{3}{2} R$
Distance $B D = 2 x = 3 R$

Q. A body performs SHM along the straight line segment ABCDE with C as the mid point of segment AE(A and E are the extreme position for the SHM). Its kinetic energies at B and D are each one fourth of its maximum value. If length of segment AE is 2R, then the distance between B and D is

23​​R

2​R​

3​R

2​R

When the particle crosses D, its speed is half the maximum speed. ∴V=Rvmax​​R2−x2​ or 2vmax​​=Rvmax​​R2−x2​ or x=23​​R Distance BD=2x=3​R

$\frac{3}{2} R$

$\frac{R}{2}$

$3 R$

$2 R$

When the particle crosses $D$ , its speed is half the maximum speed.

$∴ V = \frac{v _{m a x}}{R} R^{2} - x^{2}$
or $\frac{v _{m a x}}{2} = \frac{v _{m a x}}{R} R^{2} - x^{2}$ or $x = \frac{3}{2} R$
Distance $B D = 2 x = 3 R$