plz try to do all these questions Pakka I will mark u brain list I need all questions with step by step answer if one question is wrong then I will report it so try do this only 5question I will give u 15 points
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plz try to do it
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7
Q1 . SHM
Displacement x (t) = A Sin ωt
velocity v(t) = A ω Cos ωt
Acceleration a(t) = - A ω² Sin ωt
Given at time t, x(t) = 5m = A Sin ωt ---(1)
and | a(t) | = 20 m/s² = Aω² Sin ωt ---(2)
(2) ÷ (1) => ω² = 20/5 = 4. ω = 2 rad/sec.
f = ω/2π = 1/π rad/sec
T = 1/f = π sec.
==============
Q2. SHM
Displacement from mean position = y(t) = 24 Sin 2πt
(General form is y = A Sin ωt )
Angular frequency = ω = 2π rad/sec
frequency f = ω/2π = 1 Hz
time period = 1/f = 1 sec.
===============
Q3. SHM
v(t) = 0.007ω Cos ωt millimetres , A = 7 mm given
Maximum velocity = 4.4 m/s given.
So 0.007 ω = 4.4
ω = 4.4/0.007 rad/s
T = 2π/ω = 0.01 sec
==================
Q4. SHM
x = 5 cos(2πt + π/4) meters
at t = 1.5 s, x(t=1.5s) = 5* Cos (3π+π/4) = - 5/√2 meters
v(t) = - 5 * 2π * Sin (2π t + π/4) m/sec
So v(t =1.5 s) = - 5* 2π * sin (3π+π/4) = + 5 π √2 m/s
a(t) = - 5 * (2π)² Cos(2πt + π/4) m/s²
a(1.5 s) = + 10 π² √2 m/s²
===============
Q5. Simple pendulum.
Given that its time period T = 2 second. (from one side to another it is 1 sec).
Time period = 2 π √(L/g)
L = g T² / (4 π²)
= 9.81 * 2² / (4 π²)
= 0.994 m ≈ 1 m
Displacement x (t) = A Sin ωt
velocity v(t) = A ω Cos ωt
Acceleration a(t) = - A ω² Sin ωt
Given at time t, x(t) = 5m = A Sin ωt ---(1)
and | a(t) | = 20 m/s² = Aω² Sin ωt ---(2)
(2) ÷ (1) => ω² = 20/5 = 4. ω = 2 rad/sec.
f = ω/2π = 1/π rad/sec
T = 1/f = π sec.
==============
Q2. SHM
Displacement from mean position = y(t) = 24 Sin 2πt
(General form is y = A Sin ωt )
Angular frequency = ω = 2π rad/sec
frequency f = ω/2π = 1 Hz
time period = 1/f = 1 sec.
===============
Q3. SHM
v(t) = 0.007ω Cos ωt millimetres , A = 7 mm given
Maximum velocity = 4.4 m/s given.
So 0.007 ω = 4.4
ω = 4.4/0.007 rad/s
T = 2π/ω = 0.01 sec
==================
Q4. SHM
x = 5 cos(2πt + π/4) meters
at t = 1.5 s, x(t=1.5s) = 5* Cos (3π+π/4) = - 5/√2 meters
v(t) = - 5 * 2π * Sin (2π t + π/4) m/sec
So v(t =1.5 s) = - 5* 2π * sin (3π+π/4) = + 5 π √2 m/s
a(t) = - 5 * (2π)² Cos(2πt + π/4) m/s²
a(1.5 s) = + 10 π² √2 m/s²
===============
Q5. Simple pendulum.
Given that its time period T = 2 second. (from one side to another it is 1 sec).
Time period = 2 π √(L/g)
L = g T² / (4 π²)
= 9.81 * 2² / (4 π²)
= 0.994 m ≈ 1 m
Answered by
0
the correct answer is 1 metre
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