A body executing linear S.H.M has a velocity of 3cm/s. When its displacement is 4cm. And a velocity of 4 cm/s. When it's displacement is 3
a) find amplitude and period of oscillation
b) if the mass of the body is 100 g. Calculate total energy of oscillation.
Answers
Hello buddy,
◆ Answer-
a = 5 cm
T = 6.284 s
E = 1.25×10^-4 J
◆ Explaination-
# Given-
v1 = 3 cm/s at x1 = 4 cm
v2 = 4 cm/s at x2 = 3 cm
m = 100 g = 0.1 kg
# Solution-
Velocity of the particle in SHM is given by-
v = ω √(a^2-x^2)
v^2 = ω^2 (a^2-x^2)
Therefore,
v1^2 = ω^2 (a^2-x1^2)
3^2 = ω^2 (a^2-4^2)
9 = ω^2 (a^2-16) ...(1)
v2^2 = ω^2 (a^2-x2^2)
4^2 = ω^2 (a^2-3^2)
16 = ω^2 (a^2-9) ...(2)
Solving (1) & (2), you'll get
a = 5 cm
ω = 1 rad/s
Period of oscillation is calculated by-
T = 2π/ω
T = 2×3.142/1
T = 6.284 s
Total energy in oscillation is calculated by-
E = 1/2 m(ωa)^2
E = 1/2 × 0.1 × (1×0.05)^2
E = 1.25×10^-4 J
Hope this helps...
Hello buddy,
◆ Answer-
a = 5 cm
T = 6.284 s
E = 1.25×10^-4 J
◆ Explaination-
# Given-
v1 = 3 cm/s at x1 = 4 cm
v2 = 4 cm/s at x2 = 3 cm
m = 100 g = 0.1 kg
# Solution-
Velocity of the particle in SHM is given by-
v = ω √(a^2-x^2)
v^2 = ω^2 (a^2-x^2)
Therefore,
v1^2 = ω^2 (a^2-x1^2)
3^2 = ω^2 (a^2-4^2)
9 = ω^2 (a^2-16) ...(1)
v2^2 = ω^2 (a^2-x2^2)
4^2 = ω^2 (a^2-3^2)
16 = ω^2 (a^2-9) ...(2)
Solving (1) & (2), you'll get
a = 5 cm
ω = 1 rad/s
Period of oscillation is calculated by-
T = 2π/ω
T = 2×3.142/1
T = 6.284 s
Total energy in oscillation is calculated by-
E = 1/2 m(ωa)^2
E = 1/2 × 0.1 × (1×0.05)^2
E = 1.25×10^-4 J
Hope this helps...