A particle executes SHM of amplitude 25cm and time period 3s. What is the minimum time required for the particle to move between two points located at 12.5cm on either side of the mean position.
A
0.8s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.5s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.3s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.0s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B0.5s
Let C and D be the two extreme position of the particle, with O as mean position.
Displacements to the right of O are taken as positive while those to the left of O are taken as negative.
The displacement at any instant of a particle executing SHM is given by
x(t)=Asin(ωt+ϕ)
Here, x(t)=25sin(2π3t+ϕ)
Let's suppose that at t=0, the particle is at point D.
So, x(0)=25=25sinϕ
⇒ϕ=π2
∴x(t)=25sin(2π3t+π2)=25cos(2π3t)
Now, assume that particle reaches at A at t=t1 and at B at t=t2.