已知某质点作简谐振动，其振动方程为=0.6cos (5pi t+0.8pi ),则初相位等于___=0.6cos (5pi t+0.8pi )。

已知简谐运动方程为=0.6cos (dfrac (pi t)(4)+dfrac (5pi )(4)),则振动周期为( )s.已知简谐运动方程为,则振动周期为()

,两个同方向同频率的简谐振动,其振动表达式分别_(1)=0.6cos (5t+dfrac (pi )(2))(m),_(1)=0.6cos (5t+dfrac

已知一质点作简谐振动振动方程为=0.02cos (50pi t-dfrac (pi )(3)),则该质点在t=2s时的速度v=( )=0.02cos (5

(10)一质点做简谐振动,其振动方程为-|||-=3times (10)^-2cos (5pi t+0.25pi ) (SI单位)-|||-求振动的振幅、频率和

一个作简谐振动的质点，其运动方程为x=5cos(({pi )over(2) }t+({pi )over(2) } )(cm)，则振动的振幅A= ,

计算题题号:分值:10分难度系数品级:1一质量为0.20 kg的质点作简谐振动,其振动方程为=0.6cos (5t+dfrac (1)(2)pi )-|||-_

已知一质点作简谐振动振动方程为=Asin (omega t+dfrac (pi )(3))则其初相位为=Asin (omega t+dfrac (pi )(3)

【题目】某质点参与 _(1)=5cos (pi t-dfrac (pi )(2))cm 及-|||-_(2)=5cos (pi t+dfrac (pi )(2)

两简谐振动的振动方程分别为_(1)=3cos (2pi t-dfrac (pi )(3)), _(2)=4cos (2pi t+dfrac (pi )(6))单

某质点参与_(1)=10cos (pi t-dfrac (pi )(2))cm 及 _(2)=20cos (pi t+dfrac (pi )(2))cm两个同方