A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ , respectively. Its time period is:
$2 π \sqrt{\frac{V_{1}^{2} + V_{2}^{2}}{x_{1}^{2} + x_{2}^{2}}}$
$2 π \sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{x_{1}^{2} - x_{2}^{2}}}$
$2 π \sqrt{\frac{x_{1}^{2} + x_{2}^{2}}{V_{1}^{2} + V_{2}^{2}}}$
$2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$

Question

A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ , respectively. Its time period is:
$2 π \sqrt{\frac{V_{1}^{2} + V_{2}^{2}}{x_{1}^{2} + x_{2}^{2}}}$
$2 π \sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{x_{1}^{2} - x_{2}^{2}}}$
$2 π \sqrt{\frac{x_{1}^{2} + x_{2}^{2}}{V_{1}^{2} + V_{2}^{2}}}$
$2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$

Toppr Admin · Accepted Answer

-Using the-expressions for velocity for SHM-v1-x3C9-x221A-A2-x2212-x21v2-x3C9-x221A-A2-x2212-x22Squaring-both-v21-x3C9-2-A2-x2212-x21- -3-v22-x3C9-2-A2-x2212-x22- -4-Subtracting -4- from -3- v21-x2212-v22-x3C9-2-x22-x2212-x21-x21D2-4-x3C0-2-x22-x2212-x21-T2-v21-x2212-v22-x21D2-T-2-x3C0-xE001-xE000-xE000-x23B7-x22-x2212-x21v21-x2212-v22 where we have used -x3C9-2-x3C0-T-

A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is:2π√V21+V22x21+x222π√V21−V22x21−x222π√x21+x22V21+V222π√x22−x21V21−V22

Using the expressions for velocity for SHM,v1=ω√A2−x21v2=ω√A2−x22Squaring both,v21=ω2(A2−x21) ....(3)v22=ω2(A2−x22) ....(4)Subtracting (4) from (3), v21−v22=ω2(x22−x21)⇒4π2(x22−x21)=T2(v21−v22)⇒T=2π ⎷x22−x21v21−v22 where we have used ω=2πT