A particle executes simple harmonic motion between x = -A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then?
A.
T1 < T2
B.
T1 > T2
C.
T1=T2
D.
T1=2T2
Answer: Option A
Explanation:
x=Asinωt For x=A/2, sinωT1=1/2 ωT1=π/6 or T1=π/2ω For x=A, sinω(T1+T2)=1 T1+T2=π/2ω T2=π/2ω-π/6ω=π/3ω=2T1.
A simple pendulum has a time period T1, when on the earth’s surface; and T2, when taken to a height R above the earth’s surface (R is the radius of the earth). The value of T2/T1 is?
A.
2
B.
√2
C.
4
D.
1
Answer: Option D
Explanation:
g‘/g=(R/(R+R))2=1/4 As T∝1/√g T2/T1 = √(g/g‘)=√(4/1)=2.
A child swinging on a swing stands up. Then the time period of the swing will?
A.
Increase
B.
Decrease
C.
Remain the same
D.
Increase, if the child is long and decreases if the child is short
Answer: Option B
Explanation:
The child and the swing together constitute a pendulum of a time period, T∝2π√(l/g) As the girl stands up, her centre of gravity is raised. The distance between the point of suspension and the centre of gravity decreases, that is the length l decreases. Hence the time period T decreases.
