The displacement of a particle executing SHM at any instant t is $x = 7 cos 0.5 π t$ . The time taken by the particle in reaching from mean position to extreme position will be

Question

The displacement of a particle executing SHM any instant t is x-7cos-0-5 pi t- The time taken by the particle in reaching from mean position to extreme position will be4 s2 s1 s1-2 s

Toppr Admin · Accepted Answer

It takes T4 seconds to go from mean to extreme position -2-x3C0-xA0-in-xA0-T-xA0-means-xA0-x3C0-2-xA0-in-xA0-T4-Here- -x3C9-0-5-x3C0- thus T-2-x3C0-x3C9-4Thus-xA0-T4-1-xA0-s

The displacement of a particle executing SHM at any instant t is $x = 7 cos 0.5 π t$ . The time taken by the particle in reaching from mean position to extreme position will be
$4 s$
$2 s$
$1 s$
$1 / 2 s$

It takes $\frac{T}{4}$ seconds to go from mean to extreme position $(2 π in T, means \frac{π}{2} in \frac{T}{4})$
Here, $ω = 0.5 π$ , thus $T = \frac{2 π}{ω} = 4$
Thus $, \frac{T}{4} = 1 s$

The displacement of a particle executing SHM at any instant t is x=7cos0.5πt. The time taken by the particle in reaching from mean position to extreme position will be4s2s1s1/2s

It takes T4 seconds to go from mean to extreme position (2π in T, means π2 in T4)Here, ω=0.5π, thus T=2πω=4Thus, T4=1 s

The displacement of a particle executing SHM at any instant t is $x = 7 cos 0.5 π t$ . The time taken by the particle in reaching from mean position to extreme position will be
$4 s$
$2 s$
$1 s$
$1 / 2 s$

It takes $\frac{T}{4}$ seconds to go from mean to extreme position $(2 π in T, means \frac{π}{2} in \frac{T}{4})$
Here, $ω = 0.5 π$ , thus $T = \frac{2 π}{ω} = 4$
Thus $, \frac{T}{4} = 1 s$